Chapter 9.4, Problem 22E

To determine

Prove the formula for all integers $n\ge 1$ .

Answer to Problem 22E

  $\mathrm{Pr}oved$

Explanation of Solution

Calculation:

Consider the given equation,

  ${\displaystyle \sum _{i=1}^n{i}^4=\frac{n\left(n+1\right)\left(2n+1\right)\left(3n^2+3n-1\right)}{30}}$

Now prove the formula for $n=1$

  $\begin{array}{l}{S}_n={\displaystyle \sum _{i=1}^n{i}^4}\\ =\frac{n\left(n+1\right)\left(2n+1\right)\left(3n^2+3n-1\right)}{30}\\ put,n=1\\ S_1=\frac{1\left(1+1\right)\left(2\times 1+1\right)\left(3{\left(1\right)}^2+3\times 1-1\right)}{30}\\ S_1=\frac{30}{30}=1\end{array}$

This is valid,

Now prove for second part, assume that $n=k$ ,

  $\begin{array}{l}{S}_k={\displaystyle \sum _{i=1}^k{i}^4}\\ S_k=\frac{k\left(k+1\right)\left(2k+1\right)\left(3k^2+3k-1\right)}{30}\end{array}$

Now assume $n=k+1$ ,

  $S_{k+1}={\displaystyle \sum _{i=1}^{k+1}{i}^4}+{\left(k+1\right)}^4$

  $\begin{array}{l}{S}_{k+1}=\frac{k\left(k+1\right)\left(2k+1\right)\left(3k^2+3k-1\right)}{30}+{\left(k+1\right)}^4\\ =\frac{k\left(k+1\right)\left(2k+1\right)\left(3k^2+3k-1\right)30{\left(k+1\right)}^4}{30}\\ =\frac{\left(k+1\right)\left(k\left(k+1\right)\left(2k+1\right)\left(3k^2+3k-1\right)30{\left(k+1\right)}^3\right)}{30}\\ =\frac{\left(k+1\right)\left(\left(2k^2+k\right)\left(3k^2+3k-1\right)+30\left(k^3+3k^2+3k+1\right)\right)}{30}\\ =\frac{\left(k+1\right)\left(\left(6k^4+6k^3-2k^2+3k^3+3k^2-k+30k^3+90k^2+90k+30\right)\right)}{30}\\ =\frac{\left(k+1\right)\left(6k^4+39k^3+91k^2+89k+30\right)}{30}\\ =\frac{\left(k+1\right)\left(2k^2+7k+6\right)\left(3k^2+9k+5\right)}{30}\\ =\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)\left(3k^2+9k+5\right)}{30}\end{array}$

On combining both the results we conclude that the formula is valid for all positive integer values of $n$ .

Hence, the result proved.