Question

Verify that $\sum _{k=1}^n{k}^2\mathrm{is}\mathrm{equal}\mathrm{to}\frac{n(n+1)(2n+1)}{6}$for the cases $\left(a\right)n=1,\left(b\right)n=5,\left(c\right)n=10$

Short answer

$$\begin{gathered} \mathrm{Part}\left(\mathrm{a}\right)Forn=1,\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6} \\ \mathrm{Part}\left(b\right)Forn=5,\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6} \\ \mathrm{Part}\left(c\right)Forn=10,\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6} \end{gathered}$$

Step-by-step solution

Part(a) Step 1. Given Information   

The given sum is$\sum _{k=1}^n{k}^2$

Part(a) Step 2. Calculation   

The left-hand side sum and the right-hand side sum is as follows,

$$\begin{gathered} \sum _{k=1}^1{k}^2=1 \\ And \\ \frac{n(n+1)(2n+1)}{6}=\frac{1(1+1)\left(2\right(1)+1)}{6} \\ =\frac{6}{6} \\ =1 \end{gathered}$$

Thus, it is verified.

Part(c) Step 1. Calculation   

The left-hand side sum and the right-hand side sum are as follows,

$$\begin{gathered} \sum _{k=1}^5{k}^2=1^2+2^2+3^2+4^2+5^2 \\ =55 \\ And \\ \frac{n(n+1)(2n+1)}{6}=\frac{5(5+1)\left(2\right(5)+1)}{6} \\ =\frac{5\times 6\times 11}{6} \\ =55 \end{gathered}$$

Thus, it is verified.

Part(c) Step 1. Calculation   

The left-hand side sum and the right-hand side sum are as follows,

$$\begin{gathered} \sum _{k=1}^{10}{k}^2=1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+{10}^2 \\ =385 \\ And \\ \frac{n(n+1)(2n+1)}{6}=\frac{10(10+1)\left(2\right(10)+1)}{6} \\ =\frac{10\times 11\times 21}{6} \\ =385 \end{gathered}$$

Thus, it is verified.