Question

If $(1+x+x^2+.....+x^p{)}^n=a_0+a_{1}x+a_2{x}^2+.....a_{np}{x}^{np}$
Prove that
$\left(1\right)a_0+a_1+a_2+.....+a_{np}=(p+1{)}^n$.
$\left(2\right)a_1+2a_2+3a_3+....+np.a_{np}=\frac{1}{2}np(p+1{)}^n$.

Answer 1

$\left\{1\right\}$

for $1st$ part we need to put $x=1$ both sides.

so $RHS$ will be $a_0+a_1+a_2+\dots a_{np}$ and $LHS$ will become equal to $\{1+1+1+\dots \{p+1\}times{\}}^n$ or $\{p+1{\}}^n$ and since $RHS=LHS$ therefore

$a_0+a_1+a_2+\dots a_{np}=\{p+1{\}}^n$

$\left\{2\right\}$

for $2nd$ part we need to differentitate the given equation and then put $x=1$.

so $RHS$ will be $a_1+2a_2+3a_3+\dots np.a_{np}$ and $LHS$ will be $n\times \{1+1+1+\dots \{p+1\}times{\}}^{n-1}\times \{1+2+3+\dots p\}$ and since sum of first $n$ natural numbers is $\frac{n\{n+1\}}{2}$ so $\{1+2+3+\dots p\}$ is equal to $\frac{p\{p+1\}}{2}$ and therefore $LHS$ is equal to $\frac{np\{p+1\}}{2}\times \{p+1{\}}^{n-1}$ or $\frac{1}{2}np\{p+1{\}}^n$.

And since $RHS=LHS$ therefore

$a_1+2a_2+3a_3+\cdots +np.a_{n}p=\frac{1}{2}np\{p+1{\}}^n$.