Question

AS WRITE ALL DETAILS AND STEPS FOR FULL CREDITS 1) Prove that (": +^2) = :0 () 6,2,), where n, nz,r and k are natural numbers.

Answer 1

Consider the Binomial expansions

$\fn_jvn (1+x)^{n_1}=\binom{n_1}{0}+\binom{n_1}{1}x+\binom{n_1}{2}x^2+....+\binom{n_1}{n_1}x^{n_1}$

$\fn_jvn (1+x)^{n_2}=\binom{n_2}{0}+\binom{n_2}{1}x+\binom{n_2}{2}x^2+....+\binom{n_2}{n_2}x^{n_2}$

Multiplying the above expansions,

$\fn_jvn (1+x)^{n_1}(1+x)^{n_2}=(1+x)^{n_1+n_2}=\left [\binom{n_1}{0}+\binom{n_1}{1}x+\binom{n_1}{2}x^2+....+\binom{n_1}{n_1}x^{n_1} \right ]\left [ \binom{n_2}{0}+\binom{n_2}{1}x+\binom{n_2}{2}x^2+....+\binom{n_2}{n_2}x^{n_2} \right ]$The coefficient of  $\fn_jvn x^k$ in the product on the right hand side of the equation is

$\fn_jvn \sum_{r=0}^{k}\binom{n_1}{r}\binom{n_2}{k-r}$ because $\fn_jvn \binom{n_1}{r}x^r\binom{n_2}{k-r}x^{k-r}=\binom{n_1}{r}\binom{n_2}{k-r}x^{k}$ .

The coefficient of  $\fn_jvn x^k$ in $\fn_jvn (1+x)^{n_1+n_2}$   (on the left hand side of the equation) is $\fn_jvn \binom{n_1+n_2}{k}$ .

Because of the equality,

$\fn_jvn \binom{n_1+n_2}{k}=\sum_{r=0}^{k}\binom{n_1}{r}\binom{n_2}{k-r}$

The proof is complete.