# Binomial Theorem – Divisibility and Differentiability Problems

## Binomial Theorem – Divisibility and Differentiability Problems

Divisibility: In the expansion $${{(1+\alpha )}^{n}}{{+}^{n}}{{C}_{1}}\alpha {{+}^{n}}{{C}_{2}}{{\alpha }^{2}}+…{{+}^{n}}{{C}_{n}}{{\alpha }^{n}}$$.

We can conclude that,

(i) $${{(1+\alpha )}^{n}}-1{{=}^{n}}{{C}_{1}}\alpha {{+}^{n}}{{C}_{2}}{{\alpha }^{2}}+…{{+}^{n}}{{C}_{n}}{{\alpha }^{n}}$$ is divisible by α i.e., it is a multiple of α.

Example: For all n ϵ R, 9n⁺¹ – 8n – 9 is divisible by?

Solution: Given that 9n⁺¹ – 8n – 9

= (1 + 8)n x 9 – 8n – 9

$$=\left( ^{n}{{C}_{0}}{{+}^{n}}{{C}_{1}}8{{+}^{n}}{{C}_{2}}{{8}^{2}}+….{{+}^{n}}{{C}_{n}}{{8}^{n}} \right)9-8n-9$$.

$$=\left( 1+8n{{+}^{n}}{{C}_{2}}{{8}^{2}}+….{{+}^{n}}{{C}_{n}}{{8}^{n}} \right)9-8n-9$$.

$$=9+72n+\left( ^{n}{{C}_{2}}{{8}^{2}}+….{{+}^{n}}{{C}_{n}}{{8}^{n}} \right)9-8n-9$$.

$$=\left( ^{n}{{C}_{2}}{{8}^{2}}+….{{+}^{n}}{{C}_{n}}{{8}^{n}} \right)9+64n$$.

$$=64\left( ^{n}{{C}_{2}}{{+}^{n}}{{C}_{3}}\times 8….{{+}^{n}}{{C}_{n}}{{8}^{n-2}} \right)9+64n$$.

$$=64\left[ \left( ^{n}{{C}_{2}}{{+}^{n}}{{C}_{3}}\times 8….{{+}^{n}}{{C}_{n}}{{8}^{n-2}} \right)9+n \right]$$.

= 64 x Some constant numbers = divisible by 64

Differentiability: Sometimes to generalize the result we use the differentiation

Example: series $${{(1+x)}^{2}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+…{{+}^{n}}{{C}_{n}}{{x}^{n}}$$  is equal to ?

Solution:  Given that $${{(1+x)}^{2}}{{=}^{n}}{{C}_{0}}{{+}^{n}}{{C}_{1}}x{{+}^{n}}{{C}_{2}}{{x}^{2}}+…{{+}^{n}}{{C}_{n}}{{x}^{n}}$$.

On differentiation with respect to the x

$$n{{(1+x)}^{n-1}}=0{{+}^{n}}{{C}_{1}}\times 1{{+}^{n}}{{C}_{2}}\times 2x+…$$.

Put x = 1 we get

$$n{{(1+1)}^{n-1}}=0{{+}^{n}}{{C}_{1}}\times 1{{+}^{n}}{{C}_{2}}\times 2(1)+…$$.

$$n{{(2)}^{n-1}}=0{{+}^{n}}{{C}_{1}}+2{{\ }^{n}}{{C}_{2}}+…$$.