Question

Find the sum of $1^4+2^4+3^4+4^4+\cdots$ to $n$ terms.

Answer 1

$S=1^4+2^4+3^4+4^4+...+n^4$

Using binomial theorem ,

${(x+1)}^5=x^5+5x^4+10x^3+10x^2+5x+1$

Let $x=1$ then ,

${(1+1)}^5=1^5+5\cdot 1^4+10\cdot 1^3+10\cdot 1^2+5\cdot 1+1$

When $x=2$

${(2+1)}^5=2^5+5\cdot 2^4+10\cdot 2^3+10\cdot 2^2+5\cdot 2+1$

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When $x=n$

${(n+1)}^5=n^5+5\cdot n^4+10\cdot n^3+10\cdot n^2+5\cdot n+1$

Adding all the term we get ,

$2^5+3^5+4^5+5^5+...+{(n+1)}^5=(1^5+2^5+3^5+4^5+5^5+...+n^5)+5(1^4+2^4+...+n^4)+10(1^3+2^3+...+n^3)$

$+10(1^2+2^2+3^2+...+n^2)+5(1+2+3+4+...+n)+(1+1+1+...+1)$

${(n+1)}^5-1=5S+10{\left(\frac{n(n+1)}{2}\right)}^2+10\left(\frac{n(n+1)(2n+1)}{6}\right)+5\left(\frac{n(n+1)}{2}\right)+n$

As , sum of square of $n$ natural number will be$\left(\frac{n(n+1)(2n+1)}{6}\right)$

Sum of cube of $n$ natural number will be ${\left(\frac{n(n+1)}{2}\right)}^2$

sum of $n$ natural number will be $\left(\frac{n(n+1)}{2}\right)$

$5S=-{(n+1)}^5+1+10{\left(\frac{n(n+1)}{2}\right)}^2+10\left(\frac{n(n+1)(2n+1)}{6}\right)+5\left(\frac{n(n+1)}{2}\right)+n$

$5S=-(n^5+5n^4+10n^3+10n^2+5n+1)+1+\frac{5}{2}(n^4+2n^3+n^2)+\frac{5}{3}(2n^3+3n^2+n)+\frac{5}{2}(n^2+n)+n$

$5S=\frac{6n^5+15n^4+10n^3-n}{6}$

$S=\frac{6n^5+15n^4+10n^3-n}{30}$

$S=\left(\frac{n(6n^4+15n^3+10n^2-1)}{30}\right)$

$S=\frac{n}{30}(n+1)(2n+1)(3n^2+3n-1)$