Chapter 9.5, Problem 12E

Solution

To state : the statements $P_1$ , $P_k$ and $P_{k+1}$ for the statement $P_n:{\displaystyle \sum _{k=1}^n{k}^4}=\frac{n\left(n+1\right)\left(2n+1\right)\left(3n^2+3n-1\right)}{30}$ .

The statements $P_1$ , $P_k$ and $P_{k+1}$ for the statement $P_n:{\displaystyle \sum _{k=1}^n{k}^4}=\frac{n\left(n+1\right)\left(2n+1\right)\left(3n^2+3n-1\right)}{30}$ are $P_1:1^4=\frac{1\cdot 2\cdot 3\cdot 5}{30}$ , $P_k:{\displaystyle \sum _{k=1}^n{k}^4}=\frac{k\left(k+1\right)\left(2k+1\right)\left(3k^2+3k-1\right)}{30}$ and $P_{k+1}:{\displaystyle \sum _{i=1}^{k+1}{k}^4}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)\left(3k^2+9k+5\right)}{30}$ .

Given information :

The statement $P_n:{\displaystyle \sum _{k=1}^n{k}^4}=\frac{n\left(n+1\right)\left(2n+1\right)\left(3n^2+3n-1\right)}{30}$ .

Explanation :

Consider the statement $P_n:{\displaystyle \sum _{k=1}^n{k}^4}=\frac{n\left(n+1\right)\left(2n+1\right)\left(3n^2+3n-1\right)}{30}$ ,

Write $P_1$ by replacing $n$ with 1.

  $P_1:{\displaystyle \sum _{k=1}^n{k}^4}=\frac{1\left(1+1\right)\left(2\cdot 1+1\right)\left(3\cdot 1^2+3\cdot 1-1\right)}{30}$

  $P_1:1^4=\frac{1\cdot 2\cdot 3\cdot 5}{30}$

Write $P_k$ by replacing $n$ with $k$ . Also replace $i$ with $k$ in the summation.

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

Write $P_{k+1}$ by replacing $n$ with $k+1$ in $P_k$ .

  $P_{k+1}:{\displaystyle \sum _{i=1}^{k+1}{k}^4}=\frac{\left(k+1\right)\left[\left(k+1\right)+1\right]\left[2\left(k+1\right)+1\right]\left[3{\left(k+1\right)}^2+3\left(k+1\right)-1\right]}{30}$

  $P_{k+1}:{\displaystyle \sum _{i=1}^{k+1}{k}^4}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)\left(3k^2+6k+3+3k+3-1\right)}{30}$

  $P_{k+1}:{\displaystyle \sum _{i=1}^{k+1}{k}^4}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)\left(3k^2+9k+5\right)}{30}$

Thus, the statements $P_1$ , $P_k$ and $P_{k+1}$ for the statement $P_n:{\displaystyle \sum _{k=1}^n{k}^4}=\frac{n\left(n+1\right)\left(2n+1\right)\left(3n^2+3n-1\right)}{30}$ and $P_1:1^4=\frac{1\cdot 2\cdot 3\cdot 5}{30}$ , $P_k:{\displaystyle \sum _{k=1}^n{k}^4}=\frac{k\left(k+1\right)\left(2k+1\right)\left(3k^2+3k-1\right)}{30}$ and $P_{k+1}:{\displaystyle \sum _{i=1}^{k+1}{k}^4}=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)\left(3k^2+9k+5\right)}{30}$ .