Question

If $a,b,p\in N$ and $p$ is prime, then ${(a+b)}^p-a^p-b^p$ is divisible by

• A. $p$
• B. $\frac{1}{2}p(p-1)$
• C. $p^2$
• D. None of these

Answer 1

Answer: (A)

$p$

$(a+b{)}^p-a^p-b^p$

$=[a^p+{}^p{C}_1{a}^{p-1}b+...+{}^p{C}_{p-1}a{b}^{p-1}+{}^p{C}_p{a}^0{b}^p]-a^p-b^p$
$={}^p{C}_1{a}^{p-1}b+...+{}^p{C}_{p-1}a{b}^{p-1}$

$=p\left(a^{p-1}b\right)+\frac{p(p-1)}{2}{a}^{p-2}{b}^2+...+pa{b}^{p-1}$

$=p\left(\right(a^{p-1}b)+\frac{(p-1)}{2}{a}^{p-2}{b}^2+...+a{b}^{p-1})$

Hence the above expression is divisible by $p$.