Applications of Binomial Theorem

Introduction

The Binomial Theorem is not just for expanding expressions; it is a powerful tool used in various mathematical applications, particularly in number theory and analysis.

1. Divisibility and Remainder Problems

This is one of the most common applications of the binomial theorem. The core idea is to express a number in a binomial form related to the divisor.

• Technique: To find the remainder when $a^n$ is divided by $d$, express 'a' in the form $(kd+b)$ or $(kd-b)$, where b is as small as possible. Then expand using the theorem.

$$a^n = (kd+b)^n = {^nC_0}(kd)^n + ... + {^nC_{n-1}}(kd)^1b^{n-1} + {^nC_n}b^n$$

All terms except the last one ($b^n$) are multiples of $d$. Therefore, the remainder is the same as the remainder when $b^n$ is divided by $d$.

2. Approximations

For a rational index n, the binomial expansion $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + ...$ can be used to find approximate values if $|x|$ is very small. For a quick approximation, we can use the first two terms:

$$(1+x)^n \approx 1 + nx \quad (\text{for } |x| \ll 1)$$

This is useful for calculating roots and powers of numbers that are close to 1.

3. Summation of Series

Many complex-looking series involving binomial coefficients ($^nC_r$) can be evaluated by differentiating or integrating the binomial expansion of $(1+x)^n$ and then substituting a suitable value for x.

• Differentiation: Differentiating $\sum {^nC_r}x^r = (1+x)^n$ can help solve series involving a factor of 'r'.
• Integration: Integrating $\sum {^nC_r}x^r = (1+x)^n$ can help solve series where $^nC_r$ is divided by a factor like 'r+1'.

Example 1: Divisibility

Question: Show that $9^{n+1} - 8n - 9$ is divisible by 64 for any positive integer n.

Solution: We write $9^{n+1} = (1+8)^{n+1}$. Using the Binomial Theorem:

$$(1+8)^{n+1} = {^{n+1}C_0}(1)^{n+1} + {^{n+1}C_1}(1)^n(8)^1 + {^{n+1}C_2}(8)^2 + {^{n+1}C_3}(8)^3 + ...$$

$$= 1 + (n+1)8 + {^{n+1}C_2}(64) + {^{n+1}C_3}(512) + ...$$

$$9^{n+1} = 1 + 8n + 8 + 64 \cdot {^{n+1}C_2} + ...$$

$$9^{n+1} = 9 + 8n + 64 [ {^{n+1}C_2} + {^{n+1}C_3}(8) + ... ]$$

$$9^{n+1} - 8n - 9 = 64 \times (\text{an integer})$$

Since the expression is a multiple of 64, it is divisible by 64.

Example 2: Finding Remainders

Question: Find the remainder when $5^{99}$ is divided by 13.

Solution: We need to express a power of 5 in terms of a multiple of 13. We know $5^2 = 25 = 26-1 = 2 \times 13 - 1$.

$$5^{99} = 5 \cdot 5^{98} = 5 \cdot (5^2)^{49} = 5(26-1)^{49}$$

Expanding $(26-1)^{49}$ using the binomial theorem gives:

$$= 5[ {^{49}C_0}(26)^{49} - {^{49}C_1}(26)^{48} + ... - {^{49}C_{48}}(26)^1 + {^{49}C_{49}}(-1)^{49} ]$$

Every term in the bracket except the last one is a multiple of 26, and thus a multiple of 13. Let the sum of these terms be $13k$.

$$= 5[13k + 1(-1)] = 65k - 5$$

When $65k-5$ is divided by 13, the remainder is -5. To get a positive remainder, we add the divisor: $-5 + 13 = 8$. The remainder is 8.

Example 3: Summation of Series by Differentiation

Question: Find the value of the sum $S = \sum_{r=1}^{n} r \cdot {^nC_r}$.

Solution: We start with the identity $(1+x)^n = {^nC_0} + {^nC_1}x + {^nC_2}x^2 + ... + {^nC_n}x^n$. Differentiating both sides with respect to x:

$$n(1+x)^{n-1} = {^nC_1} + 2{^nC_2}x + 3{^nC_3}x^2 + ... + n{^nC_n}x^{n-1}$$

Now, substitute $x=1$ into this equation:

$$n(1+1)^{n-1} = {^nC_1} + 2{^nC_2} + 3{^nC_3} + ... + n{^nC_n}$$

This is the required sum. Therefore, the sum is $n \cdot 2^{n-1}$.