Chapter 9.4, Problem 21E

To determine

To proof: The expression ${\displaystyle \sum _{i=1}^n{i}^5}=\frac{n^2{\left(n+1\right)}^2\left(2n^2+2n-1\right)}{12}$ with help of mathematical induction for integers $n\ge 1$ .

Explanation of Solution

Given information:

The expression ${\displaystyle \sum _{i=1}^n{i}^5}=\frac{n^2{\left(n+1\right)}^2\left(2n^2+2n-1\right)}{12}$ .

Formula used:

The steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Proof:

Consider expression ${\displaystyle \sum _{i=1}^n{i}^5}=\frac{n^2{\left(n+1\right)}^2\left(2n^2+2n-1\right)}{12}$ .

Denote the expression by $S_n$ .

Recall that the steps to prove a statement by mathematical induction are as follows:

Step 1. Show that the statement is true for $n=1$ .

Step 2. Assume that the statement is true for $n=k$ .

Step 3. Prove that the statement is true for $n=k+1$ with help of step 2.

Substitute $n=1$ in the expression,

  $\begin{array}{c}{1}^5=\frac{1^2{\left(1+1\right)}^2\left(2{\left(1\right)}^2+2\left(1\right)-1\right)}{12}\\ 1=\frac{4\left(3\right)}{12}\\ 1=1\end{array}$

The above equation holds true.

Therefore, the statement is true for $n=1$ so $S_1$ is true.

Now, assume that the statement if true for $n=k$ .

  ${\displaystyle \sum _{i=1}^k{i}^5}=\frac{k^2{\left(k+1\right)}^2\left(2k^2+2k-1\right)}{12}$

…(1)

Therefore, the statement holds true for $n=k$ so $S_k$ is true.

Now, prove it for $n=k+1$ . Write the expression for $n=k+1$ . Group the terms that involve $S_k$ .

  ${\displaystyle \sum _{i=1}^{k+1}{i}^5}={\displaystyle \sum _{i=1}^k{i}^5}+{\left(k+1\right)}^5$

Substitute the value of $S_k$ from equation (1),

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{k+1}{i}^5}={\displaystyle \sum _{i=1}^k{i}^5}+{\left(k+1\right)}^5\\ =\frac{k^2{\left(k+1\right)}^2\left(2k^2+2k-1\right)}{12}+{\left(k+1\right)}^5\\ =\frac{k^2{\left(k+1\right)}^2\left(2k^2+2k-1\right)+12{\left(k+1\right)}^5}{12}\\ =\frac{{\left(k+1\right)}^2\left[k^2\left(2k^2+2k-1\right)+12{\left(k+1\right)}^3\right]}{12}\end{array}$

Take the common terms and simplify it further as,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{k+1}{i}^5}=\frac{{\left(k+1\right)}^2\left[k^2\left(2k^2+2k-1\right)+12{\left(k+1\right)}^3\right]}{12}\\ =\frac{{\left(k+1\right)}^2\left[2k^4+2k^3-k^2+12k^3+12+36k^2+36k\right]}{12}\\ =\frac{{\left(k+1\right)}^2\left[2k^4+14k^3+35k^2+36k+12\right]}{12}\\ =\frac{{\left(k+1\right)}^2}{12}\frac{\left(k+2\right)\left[2k^4+14k^3+35k^2+36k+12\right]}{\left(k+2\right)}\end{array}$

Simplify it further as, divide $\left[2k^4+14k^3+35k^2+36k+12\right]$ by $\left(k+2\right)$ .

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{k+1}{i}^5}=\frac{{\left(k+1\right)}^2\left(k+2\right)\left(2k^3+10k^2+15k+6\right)}{12}\\ =\frac{{\left(k+1\right)}^2{\left(k+2\right)}^2\left(2k^2+6k+3\right)}{12}\\ =\frac{{\left(k+1\right)}^2{\left(k+2\right)}^2\left(2k^2+2k+4k+2+1\right)}{12}\\ =\frac{{\left(k+1\right)}^2{\left(k+2\right)}^2\left(2{\left(k+1\right)}^2+2k+1\right)}{12}\end{array}$

Rewrite the expression,

  $\begin{array}{c}{\displaystyle \sum _{i=1}^{k+1}{i}^5}=\frac{{\left(k+1\right)}^2{\left(k+2\right)}^2\left(2{\left(k+1\right)}^2+2k+1\right)}{12}\\ =\frac{{\left(k+1\right)}^2{\left(k+2\right)}^2\left(2{\left(k+1\right)}^2+2\left(k+1\right)-1\right)}{12}\end{array}$

Result obtained is that $S_k$ implies $S_{k+1}$ that is,

  ${\displaystyle \sum _{i=1}^{k+1}{i}^5}=\frac{{\left(k+1\right)}^2{\left(k+2\right)}^2\left(2{\left(k+1\right)}^2+2\left(k+1\right)-1\right)}{12}$

Hence, by the principle of mathematical induction the statement ${\displaystyle \sum _{i=1}^n{i}^5}=\frac{n^2{\left(n+1\right)}^2\left(2n^2+2n-1\right)}{12}$ is true for all integers $n\ge 1$ .