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