Question

State and prove Binomial theorem for any positive integer $n$.

Answer 1

The proof is obtained by principal of mathamatical induction

let the given statement be

$P\left(n\right):{\left(a+b\right)}^n{=}^n{C}_0{a}^n{+}^n{C}_1{a}^{n-1}b{+}^n{C}_2{a}^{n-2}{b}^2+.....{+}^n{C}_{n-1}a.b^{n-1}{+}^n{C}_n{b}^n$

for n=1

$p\left(1\right):{\left(a+b\right)}^1{=}^1{C}_0{a}^1{+}^1{C}_1{b}^1=a+b$

thus p(1) is true.

suppose p(k) is true for some positive integer k,

${\left(a+b\right)}^k{=}^k{C}_0{a}^k{+}^k{C}_1{a}^{k-1}b{+}^k{C}_2{a}^{k-2}{b}^2+.....{+}^k{C}_k{b}^k$

we shall proove that P(k+1) is also true,

${\left(a+b\right)}^{k+1}{=}^{k+1}{C}_0{a}^{k+1}{+}^{k+1}{C}_1{a}^{k}b{+}^{k+1}{C}_2{a}^{k-1}{b}^2+...{+}^{k+1}{C}_{k+1}{b}^{k+1}$

now

${\left(a+b\right)}^{k+1}=\left(a+b\right){\left(a+b\right)}^k$

$=\left(a+b\right)\left({}^k{C}_0{a}^k{+}^k{C}_1{a}^{k-1}b{+}^k{C}_2{a}^{k-2}{b}^2+...{+}^k{C}_{k-1}a{b}^{k-1}{+}^k{C}_k{b}^k\right)......\left(1\right)$

${=}^k{C}_0{a}^{k+1}{+}^k{C}_1{a}^{k}b{+}^k{C}_2{a}^{k-1}{b}^2+...{+}^k{C}_{k-1}{a}^2{b}^{k-1}{+}^k{C}_{k}a{b}^k{+}^k{C}_0{a}^{k}b{+}^k{C}_1{a}^{k-1}{b}^2{+}^k{C}_2{a}^{k-2}{b}^3$

$+...{+}^k{C}_{k-1}a{b}^k{+}^k{C}_k{b}^{k+1}$

${=}^k{C}_0{a}^{k+1}+\left({}^k{C}_1{+}^k{C}_0\right)a^{k}b+\left({}^k{C}_2{+}^k{C}_1\right)a^{k-1}{b}^2+...+\left({}^k{C}_k{+}^k{C}_{k-1}\right)a{b}^k{+}^k{C}_k{b}^{k+1}$

${=}^{k+1}{C}_0{a}^{k+1}{+}^{k+1}{C}_1{a}^{k}b{+}^{k+1}{C}_2{a}^{k-1}{b}^2+...{+}^{k+1}{C}_{k}a{b}^k{+}^{k+1}{C}_{k+1}{b}^{k+1}$

$\left(using{}^{k+1}{C}_0=1{,}^k{C}_r{+}^k{C}_{r-1}{=}^{k+1}{C}_r{,}^k{C}_k=1{=}^{k+1}{C}_{k+1}\right)$.

Thus, it has been proved that P (k + 1) is true whenever P(k) is true. Therefore, by principle of mathematical induction, P(n) is true for every positive integer n.