Pascal’s Triangle
Pascal’s triangle is a triangular arrangement of numbers in which each number (except those in the boundary) is obtained as the sum of the two numbers just above it. These numbers represent the binomial coefficients and provide a convenient way to read the coefficients in the expansion of \((a + b)^{n}\) for different positive integers \(n\).
Construction and Definition
Pascal’s triangle is built row by row. The first row consists of a single number \(1\). Each subsequent row begins and ends with \(1\), and every other entry is obtained by adding the two entries directly above it in the previous row. Symbolically, if the entry in the \(n\)th row and \(r\)th place is denoted by \({}^{n}C_{r}\), then the triangle is arranged so that the numbers in the \(n\)th row are \[ {}^{n}C_{0},\; {}^{n}C_{1},\; {}^{n}C_{2},\; \ldots,\; {}^{n}C_{n}. \] Thus, Pascal’s triangle gives a geometric arrangement of binomial coefficients.
The first few rows (written using binomial coefficients) can be described as follows. The 0th row has a single term \({}^{0}C_{0} = 1\). The 1st row contains \({}^{1}C_{0}, {}^{1}C_{1}\), the 2nd row contains \({}^{2}C_{0}, {}^{2}C_{1}, {}^{2}C_{2}\), and so on. In general, the \(n\)th row has \((n + 1)\) entries, matching exactly the number of terms in the expansion of \((a + b)^{n}\).
Derivation of Pascal’s Rule
The central relation underlying Pascal’s triangle is often called Pascal’s rule. It states that for integers \(n \ge 1\) and \(1 \le r \le n\), \[ {}^{n}C_{r} = {}^{n-1}C_{r} + {}^{n-1}C_{r-1}. \] This relation shows that each binomial coefficient in a row (except the first and last) is the sum of the two binomial coefficients directly above it, which is exactly the rule used to construct Pascal’s triangle.
To derive this identity, consider the binomial coefficients combinatorially. The number \({}^{n}C_{r}\) counts the number of ways of choosing \(r\) objects from a set of \(n\) distinct objects. Fix one particular object in this set and call it \(X\). Any selection of \(r\) objects from the \(n\) objects either includes \(X\) or does not include \(X\). If \(X\) is included, then the remaining \(r - 1\) objects must be chosen from the remaining \(n - 1\) objects, giving \({}^{n-1}C_{r-1}\) such selections. If \(X\) is not included, then all \(r\) objects must be chosen from the remaining \(n - 1\) objects, giving \({}^{n-1}C_{r}\) selections. Thus the total number of selections is the sum of these two disjoint possibilities: \[ {}^{n}C_{r} = {}^{n-1}C_{r-1} + {}^{n-1}C_{r}. \] This combinatorial argument explains the addition rule used to generate the entries of Pascal’s triangle.
Connection with the Binomial Theorem
The entries in the \(n\)th row of Pascal’s triangle are precisely the coefficients that appear in the expansion of \((a + b)^{n}\). According to the binomial theorem for positive integral indices, \[ (a + b)^{n} = \sum_{r=0}^{n} {}^{n}C_{r} a^{\,n-r} b^{\,r}. \] Here \({}^{n}C_{0}, {}^{n}C_{1}, \ldots, {}^{n}C_{n}\) are exactly the numbers found in the \(n\)th row of Pascal’s triangle. Thus, given Pascal’s triangle up to the \(n\)th row, the coefficients of \((a + b)^{n}\) can be read directly from that row without calculating each binomial coefficient separately.
For example, the coefficients in the expansion of \((a + b)^{4}\) are \({}^{4}C_{0}, {}^{4}C_{1}, {}^{4}C_{2}, {}^{4}C_{3}, {}^{4}C_{4}\). These numbers appear together in the 4th row of the triangle and so the expansion can be written down quickly as \[ (a + b)^{4} = {}^{4}C_{0}a^{4} + {}^{4}C_{1}a^{3}b + {}^{4}C_{2}a^{2}b^{2} + {}^{4}C_{3}ab^{3} + {}^{4}C_{4}b^{4}. \] In this way, Pascal’s triangle serves as a visual summary of the binomial coefficients arising in binomial expansions.
Symmetry and Other Important Aspects
One important feature of Pascal’s triangle is its symmetry. In the \(n\)th row, the entries satisfy \[ {}^{n}C_{r} = {}^{n}C_{n-r}, \] which means the row reads the same from left to right and from right to left. This mirrors the fact that choosing \(r\) objects from \(n\) objects is equivalent to leaving out \((n - r)\) objects, so the two counts must be equal. This symmetry also appears in the binomial expansion where the coefficients of the first and last terms, of the second and second last terms, and so on, are equal.
Another useful aspect is that the sum of all the entries in the \(n\)th row of Pascal’s triangle is \(2^{n}\). This follows from the binomial theorem by substituting \(a = 1\) and \(b = 1\), giving \[ (1 + 1)^{n} = 2^{n} = \sum_{r=0}^{n} {}^{n}C_{r}. \] This shows that adding all the numbers in any row of Pascal’s triangle results in a power of 2, with the exponent equal to the row index.
The alternating sum of the entries in the \(n\)th row is zero for \(n \ge 1\). This can be seen by putting \(a = 1\) and \(b = -1\) in the binomial theorem: \[ (1 - 1)^{n} = 0 = \sum_{r=0}^{n} {}^{n}C_{r} (-1)^{r}. \] This identity shows that the binomial coefficients in a fixed row, when taken with alternating signs, cancel each other out.
Pascal’s triangle also provides a clear illustration of how binomial coefficients in successive rows are related. Every internal entry in the triangle can be checked quickly using Pascal’s rule, while the outer boundary entries are always equal to 1, corresponding to \({}^{n}C_{0} = {}^{n}C_{n} = 1\). In the context of the Class XI binomial theorem, Pascal’s triangle is an important tool for understanding patterns in coefficients, verifying expansions, and building intuition about combinatorial identities.
Binomial Theorem for Positive Integral Indices
In algebra, expressions of the form \((a + b)^{n}\) are called binomial expressions, where \(a\) and \(b\) are real numbers and \(n\) is a positive integer. The binomial theorem gives a direct formula to expand \((a + b)^{n}\) without multiplying the binomial by itself \(n\) times.
Statement of the Theorem
Let \(n\) be a positive integer. Then the expansion of \((a + b)^{n}\) is given by \[ (a + b)^{n} = {}^{n}C_{0}\; a^{n} b^{0} + {}^{n}C_{1}\; a^{n-1} b^{1} + {}^{n}C_{2}\; a^{n-2} b^{2} + \cdots + {}^{n}C_{n-1}\; a^{1} b^{n-1}\; + {}^{n}C_{n}\; a^{0} b^{n}. \] Here, \({}^{n}C_{r}\) is the binomial coefficient, read as “\(n\) choose \(r\)”, and is defined for \(0 \le r \le n\) by \[ {}^{n}C_{r} = \frac{n!}{r!(n - r)!}. \] Thus \((a + b)^{n}\) expands into a sum of \((n + 1)\) terms, each term involving a product of a power of \(a\) and a power of \(b\) with suitable coefficients.
Derivation Using the Idea of Repeated Multiplication
Consider \((a + b)^{n} = (a + b)(a + b)\cdots(a + b)\), where the binomial \((a + b)\) occurs \(n\) times. To obtain any term of the expansion, one has to choose either \(a\) or \(b\) from each of the \(n\) factors and then multiply all the chosen symbols. Every term in the expansion arises in this way.
Suppose a particular term contains \(a^{n-r} b^{r}\). For this to happen, one must choose the symbol \(b\) from exactly \(r\) of the \(n\) factors and the symbol \(a\) from the remaining \(n - r\) factors. The number of different ways of choosing the \(r\) factors from which \(b\) is taken is \[ {}^{n}C_{r}. \] Since the order in which these \(r\) factors are selected does not matter for the product, each such selection gives rise to the same product \(a^{n-r} b^{r}\). Therefore, the total contribution of all such choices is \[ {}^{n}C_{r} a^{n-r} b^{r}. \]
Now, as \(r\) can take any integer value from \(0\) to \(n\), all possible terms in the expansion of \((a + b)^{n}\) are obtained by letting \(r\) vary through \(0, 1, 2, \ldots, n\). Adding all these contributions, we get \[ (a + b)^{n} = \sum_{r=0}^{n} {}^{n}C_{r} a^{\,n-r} b^{\,r}, \] which is the compact form of the binomial theorem for positive integral indices.
Proof by Mathematical Induction
The binomial theorem for positive integral indices can also be justified using the principle of mathematical induction on \(n\).
Base case: For \(n = 1\), \[ (a + b)^{1} = a + b. \] On the other hand, \[ \sum_{r=0}^{1} {}^{1}C_{r} a^{1-r} b^{r} = {}^{1}C_{0} a^{1} b^{0} + {}^{1}C_{1} a^{0} b^{1} = 1 \cdot a + 1 \cdot b = a + b. \] Thus, the formula holds for \(n = 1\).
Inductive step: Assume that for some positive integer \(k\), \[ (a + b)^{k} = \sum_{r=0}^{k} {}^{k}C_{r} a^{\,k-r} b^{\,r}. \] This is the induction hypothesis. Consider \((a + b)^{k+1}\). It can be written as \[ (a + b)^{k+1} = (a + b)^{k} (a + b). \] Using the induction hypothesis, we get \[ (a + b)^{k+1} = \left( \sum_{r=0}^{k} {}^{k}C_{r} a^{\,k-r} b^{\,r} \right) (a + b). \]
Now multiply term by term: \[ (a + b)^{k+1} = \sum_{r=0}^{k} {}^{k}C_{r} a^{\,k-r} b^{\,r} \cdot a \;+\; \sum_{r=0}^{k} {}^{k}C_{r} a^{\,k-r} b^{\,r} \cdot b. \] This simplifies to \[ (a + b)^{k+1} = \sum_{r=0}^{k} {}^{k}C_{r} a^{\,k+1-r} b^{\,r} + \sum_{r=0}^{k} {}^{k}C_{r} a^{\,k-r} b^{\,r+1}. \] In the second sum, put \(r + 1 = s\), so that \(r = s - 1\). As \(r\) runs from \(0\) to \(k\), the new index \(s\) runs from \(1\) to \(k + 1\). Thus, the second sum becomes \[ \sum_{s=1}^{k+1} {}^{k}C_{s-1} a^{\,k-(s-1)} b^{\,s} = \sum_{s=1}^{k+1} {}^{k}C_{s-1} a^{\,k+1-s} b^{\,s}. \]
Now rewrite the whole expression for \((a + b)^{k+1}\) with a common index (say \(r\)) again: \[ (a + b)^{k+1} = {}^{k}C_{0} a^{\,k+1} b^{0} + \sum_{r=1}^{k} \Big[ {}^{k}C_{r} + {}^{k}C_{r-1} \Big] a^{\,k+1-r} b^{\,r} + {}^{k}C_{k} a^{0} b^{\,k+1}. \] Using the well-known identity for binomial coefficients \[ {}^{k}C_{r} + {}^{k}C_{r-1} = {}^{k+1}C_{r}, \] we obtain \[ (a + b)^{k+1} = {}^{k+1}C_{0} a^{\,k+1} b^{0} + \sum_{r=1}^{k} {}^{k+1}C_{r} a^{\,k+1-r} b^{\,r} + {}^{k+1}C_{k+1} a^{0} b^{\,k+1}. \] Therefore, \[ (a + b)^{k+1} = \sum_{r=0}^{k+1} {}^{k+1}C_{r} a^{\,k+1-r} b^{\,r}. \]
Thus, if the binomial theorem holds for \(n = k\), then it also holds for \(n = k+1\). Since it is true for \(n = 1\), by mathematical induction it is true for every positive integer \(n\).
General Term and Important Aspects
In the expansion of \((a + b)^{n}\), the \((r+1)\)th term (counting from the left, starting with \(r = 0\)) is called the general term and is denoted by \(T_{r+1}\). From the binomial theorem, \[ (a + b)^{n} = \sum_{r=0}^{n} {}^{n}C_{r} a^{\,n-r} b^{\,r}, \] the general term is \[ T_{r+1} = {}^{n}C_{r} a^{\,n-r} b^{\,r}, \quad 0 \le r \le n. \] This expression is extremely useful in finding a particular term in the expansion without writing all the terms.
Some important observations about the expansion of \((a + b)^{n}\) for positive integral \(n\) are as follows. The total number of terms in the expansion is \(n + 1\). The powers of \(a\) decrease from \(n\) to \(0\) as we move from the first term to the last term, whereas the powers of \(b\) increase from \(0\) to \(n\). The binomial coefficients appearing in the expansion are symmetric, that is, \[ {}^{n}C_{r} = {}^{n}C_{n-r}, \] which means the coefficients of the terms equidistant from the beginning and the end of the expansion are equal.
The sum of all the binomial coefficients in the expansion of \((a + b)^{n}\) can be obtained by putting \(a = 1\) and \(b = 1\). Then \[ (1 + 1)^{n} = 2^{n} = \sum_{r=0}^{n} {}^{n}C_{r}. \] Similarly, by putting \(a = 1\) and \(b = -1\), we get \[ (1 - 1)^{n} = 0 = \sum_{r=0}^{n} (-1)^{r} {}^{n}C_{r}, \] which shows that the sum of the binomial coefficients with alternating signs is zero.
These properties of the binomial theorem for positive integral indices form the foundation for many results and applications in algebra, including finding specific coefficients, identifying middle terms when \(n\) is even or odd, and solving various combinatorial problems.
Middle Term of Binomial Expansion
In the expansion of a binomial of the form \((a + b)^{n}\), the term (or terms) that lie in the centre of the sequence of terms are called the middle term (or middle terms). Since the expansion of \((a + b)^{n}\) contains \((n + 1)\) terms in all, the position and number of middle terms depend on whether \(n\) is even or odd.
Using the binomial theorem for positive integral indices, the expansion of \((a + b)^{n}\) is \[ (a + b)^{n} = \sum_{r=0}^{n} {}^{n}C_{r} a^{\,n-r} b^{\,r}. \] The general term, also called the \((r+1)\)th term from the beginning, is \[ T_{r+1} = {}^{n}C_{r} a^{\,n-r} b^{\,r}, \quad 0 \le r \le n. \] This expression is the starting point for locating the middle term(s).
Case I: Middle Term When \(n\) is Even
If \(n\) is even, then \(n + 1\) is odd, so there is exactly one middle term in the expansion of \((a + b)^{n}\). For an odd number of terms, the middle term is the \(\dfrac{n+1+1}{2}\)th term, that is, the \(\left(\dfrac{n}{2} + 1\right)\)th term.
Let \(n = 2k\), where \(k\) is a positive integer. Then the total number of terms is \[ n + 1 = 2k + 1, \] and the unique middle term is the \((k + 1)\)th term. In terms of the general term formula, \[ T_{k+1} = {}^{2k}C_{k} a^{\,2k-k} b^{\,k} = {}^{2k}C_{k} a^{\,k} b^{\,k}. \] Hence, when \(n\) is even, the middle term of the expansion of \((a + b)^{n}\) is \[ T_{\frac{n}{2} + 1} = {}^{n}C_{\frac{n}{2}} a^{\,\frac{n}{2}} b^{\,\frac{n}{2}}. \]
Case II: Middle Terms When \(n\) is Odd
If \(n\) is odd, then \(n + 1\) is even, so there are two middle terms in the expansion of \((a + b)^{n}\). For an even number of terms, the middle two terms are the \(\dfrac{n+1}{2}\)th and \(\left(\dfrac{n+1}{2} + 1\right)\)th terms.
Let \(n = 2k + 1\), where \(k\) is a non-negative integer. Then the total number of terms is \[ n + 1 = 2k + 2, \] and the two middle positions are \[ \frac{2k+2}{2} = k + 1 \quad \text{and} \quad k + 2. \] Using the general term formula, the \((k + 1)\)th term and the \((k + 2)\)th term are \[ \begin{aligned} T_{k+1} &= {}^{2k+1}C_{k} a^{\,(2k+1)-k} b^{\,k} = {}^{2k+1}C_{k} a^{\,k+1} b^{\,k}, \\ T_{k+2} &= {}^{2k+1}C_{k+1} a^{\,(2k+1)-(k+1)} b^{\,k+1} = {}^{2k+1}C_{k+1} a^{\,k} b^{\,k+1}. \end{aligned} \] In terms of \(n\), these can be written as \[ \begin{aligned} \text{First middle term} &= T_{\frac{n+1}{2}} = {}^{n}C_{\frac{n-1}{2}} a^{\,\frac{n+1}{2}} b^{\,\frac{n-1}{2}}, \\ \text{Second middle term} &= T_{\frac{n+3}{2}} = {}^{n}C_{\frac{n+1}{2}} a^{\,\frac{n-1}{2}} b^{\,\frac{n+1}{2}}. \end{aligned} \] Thus, when \(n\) is odd, the expansion has two middle terms, given by these expressions.
Derivation of Position of Middle Term(s)
The key observation is that the expansion of \((a + b)^{n}\) has exactly \((n + 1)\) terms. To find the middle term(s), we examine the position(s) that lie in the middle of this sequence.
If the number of terms \((n + 1)\) is odd, say \(2m + 1\), then the middle term is the \((m + 1)\)th term, because there are \(m\) terms before it and \(m\) terms after it. Setting \(n + 1 = 2m + 1\) gives \(m = \dfrac{n}{2}\). Hence the middle term is the \(\left(\dfrac{n}{2} + 1\right)\)th term.
If the number of terms \((n + 1)\) is even, say \(2m\), then there is no single central term. Instead, the two middle terms are the \(m\)th and \((m + 1)\)th terms. Here \(n + 1 = 2m\), so \(m = \dfrac{n+1}{2}\). Therefore, the two middle terms are the \(\dfrac{n+1}{2}\)th and \(\left(\dfrac{n+1}{2} + 1\right)\)th terms.
Important Aspects and Observations
The middle term(s) often carry special importance in problems involving the binomial theorem. In many cases, the coefficient of the middle term (or the coefficients of the two middle terms) are the largest binomial coefficients in the corresponding expansion. This happens because the binomial coefficients \({}^{n}C_{r}\) first increase with \(r\), reach a maximum in the middle, and then decrease symmetrically.
Moreover, when expressions like \((x + \frac{1}{x})^{n}\) are expanded, the middle term(s) are frequently used to obtain terms independent of \(x\) or to locate terms with specific powers. In such problems, one typically combines the general term \[ T_{r+1} = {}^{n}C_{r} a^{\,n-r} b^{\,r} \] with the position formulas for the middle term(s) derived above to quickly identify the required term in the expansion.
Example-1
Expand \(\left(x^2+\dfrac{3}{x}\right)^4\;x\ne0\)
Solution
Here, the expression to be expanded is \(\left(x^{2} + \dfrac{3}{x}\right)^{4}\) with \(x \ne 0\). This is a binomial of the form \((a + b)^{n}\), where \(a = x^{2}\), \(b = \dfrac{3}{x}\) and \(n = 4\). Using the binomial theorem, \[ (a + b)^{n} = {}^{n}C_{0} a^{n} + {}^{n}C_{1} a^{n-1} b + {}^{n}C_{2} a^{n-2} b^{2} + \cdots + {}^{n}C_{n} b^{n}, \] we expand term by term.
\[ \begin{aligned} \left(x^{2} + \dfrac{3}{x}\right)^{4} &= {}^{4}C_{0} \left(x^{2}\right)^{4} \left(\dfrac{3}{x}\right)^{0} + {}^{4}C_{1} \left(x^{2}\right)^{3} \left(\dfrac{3}{x}\right)^{1} + {}^{4}C_{2} \left(x^{2}\right)^{2} \left(\dfrac{3}{x}\right)^{2} + {}^{4}C_{3} \left(x^{2}\right)^{1} \left(\dfrac{3}{x}\right)^{3} + {}^{4}C_{4} \left(x^{2}\right)^{0} \left(\dfrac{3}{x}\right)^{4} \\\\ &= 1 \cdot x^{8} \cdot 1 + 4 \cdot x^{6} \cdot \dfrac{3}{x} + 6 \cdot x^{4} \cdot \left(\dfrac{3}{x}\right)^{2} + 4 \cdot x^{2} \cdot \left(\dfrac{3}{x}\right)^{3} + 1 \cdot 1 \cdot \left(\dfrac{3}{x}\right)^{4} \\\\ &= x^{8} + 4x^{6} \cdot \dfrac{3}{x} + 6x^{4} \cdot \dfrac{9}{x^{2}} + 4x^{2} \cdot \dfrac{27}{x^{3}} + \dfrac{81}{x^{4}} \\\\ &= x^{8} + 12x^{5} + 54x^{2} + \dfrac{108}{x} + \dfrac{81}{x^{4}} \end{aligned} \]
Thus, the expansion of \(\left(x^{2} + \dfrac{3}{x}\right)^{4}\) for \(x \ne 0\) is \[ x^{8} + 12x^{5} + 54x^{2} + \dfrac{108}{x} + \dfrac{81}{x^{4}}. \]
Example-2
Compute \((98)^5\)
Solution
To compute \((98)^{5}\), the idea is to rewrite 98 as a number close to a power of 10 so that the binomial theorem can be applied conveniently. Since \(98 = 100 - 2\), we write \[ (98)^{5} = (100 - 2)^{5}. \] Now \((100 - 2)^{5}\) is a binomial of the form \((a + b)^{n}\) with \(a = 100\), \(b = -2\) and \(n = 5\).
Using the binomial theorem, \[ (a + b)^{n} = {}^{n}C_{0} a^{n} + {}^{n}C_{1} a^{n-1}b + {}^{n}C_{2} a^{n-2}b^{2} + \cdots + {}^{n}C_{n} b^{n}, \] we expand \((100 - 2)^{5}\) as follows.
\[ \begin{aligned} (98)^{5} &= (100 - 2)^{5} \\ &= {}^{5}C_{0} (100)^{5} ( -2 )^{0} - {}^{5}C_{1} (100)^{4} (2) + {}^{5}C_{2} (100)^{3} (2)^{2} - {}^{5}C_{3} (100)^{2} (2)^{3} + {}^{5}C_{4} (100) (2)^{4} - {}^{5}C_{5} (2)^{5} \\ &= 1 \cdot 100^{5} \cdot 1 - 5 \cdot 100^{4} \cdot 2 + 10 \cdot 100^{3} \cdot 4 - 10 \cdot 100^{2} \cdot 8 + 5 \cdot 100 \cdot 16 - 1 \cdot 32 \\ &= 10000000000 - 1000000000 + 40000000 - 800000 + 8000 - 32. \end{aligned} \]
\[ \begin{aligned} (98)^{5} &= 10000000000 - 1000000000 \\ &= 9000000000 \\ &= 9000000000 + 40000000 \\ &= 9040000000 \\ &= 9040000000 - 800000 \\ &= 9039200000 \\ &= 9039200000 + 8000 \\ &= 9039208000 \\ &= 9039208000 - 32 \\ &= 9039207968 \end{aligned} \]
Therefore, the value of \((98)^{5}\) is \(9039207968\).
Example-3
Using binomial theorem, prove that \(6^n–5n\) always leaves remainder 1 when divided by 25.
Solution
The expression to be considered is \(6^{n} - 5n\). To show that this always leaves remainder 1 when divided by 25, it is enough to prove that \(6^{n} - 5n\) can be written in the form \(25k + 1\) for some natural number \(k\). This will mean that \(6^{n} - 5n\) is exactly 1 more than a multiple of 25.
Observe that \(6 = 1 + 5\). Hence, \[ 6^{n} = (1 + 5)^{n} \] Using the binomial theorem, \[ (1 + a)^{n} = {}^{n}C_{0} \cdot 1^{n} a^{0} + {}^{n}C_{1} \cdot 1^{n-1} a^{1} + {}^{n}C_{2} \cdot 1^{n-2} a^{2} + \ldots + {}^{n}C_{n} \cdot 1^{0} a^{n} \] Taking \(a = 5\), we obtain \[ \begin{aligned} (1 + 5)^{n} &= {}^{n}C_{0} \cdot 1^{n} \cdot 5^{0} + {}^{n}C_{1} \cdot 1^{n-1} \cdot 5^{1} + {}^{n}C_{2} \cdot 1^{n-2} \cdot 5^{2} + {}^{n}C_{3} \cdot 1^{n-3} \cdot 5^{3} + \ldots + {}^{n}C_{n} \cdot 1^{0} \cdot 5^{n} \\ &= 1 + {}^{n}C_{1} \cdot 5 + {}^{n}C_{2} \cdot 5^{2} + {}^{n}C_{3} \cdot 5^{3} + \ldots + {}^{n}C_{n} \cdot 5^{n} \end{aligned} \]
Since \({}^{n}C_{1} = n\), the above expansion can be written as \[ \begin{aligned} 6^{n} &= 1 + 5n + {}^{n}C_{2} \cdot 5^{2} + {}^{n}C_{3} \cdot 5^{3} + \ldots + {}^{n}C_{n} \cdot 5^{n}. \end{aligned} \] Now subtract \(5n\) from both sides: \[ \begin{aligned} 6^{n} - 5n &= 1 + {}^{n}C_{2} \cdot 5^{2} + {}^{n}C_{3} \cdot 5^{3} + \ldots + {}^{n}C_{n} \cdot 5^{n} \end{aligned} \]
Each term from \({}^{n}C_{2} \cdot 5^{2}\) onwards contains the factor \(5^{2} = 25\). Therefore, these terms are all multiples of 25. Hence, we can factor out 25 from the right-hand side: \[ \begin{aligned} 6^{n} - 5n &= 1 + 25 \left( {}^{n}C_{2} + {}^{n}C_{3} \cdot 5 + {}^{n}C_{4} \cdot 5^{2} + \ldots + {}^{n}C_{n} \cdot 5^{\,n-2} \right) \end{aligned} \] Let \[ k = {}^{n}C_{2} + {}^{n}C_{3} \cdot 5 + {}^{n}C_{4} \cdot 5^{2} + \ldots + {}^{n}C_{n} \cdot 5^{\,n-2}, \] which is a natural number (being a sum of products of integers). Then \[ 6^{n} - 5n = 25k + 1 \]
Thus \(6^{n} - 5n\) is of the form \(25k + 1\), which shows that on division by 25, it always leaves remainder 1.