Question

Use the technique given in Exercise 35, together with the result of Exercise , to derive the formula for$\sum _{k=1}^n k^2$ given in Table 2. [Hint: Take${\mathrm{a}}_{\mathrm{k}}={\mathrm{k}}^3$ in the telescoping sum in Exercise 35.]

Short answer

Thus, the formula for,$\sum _{\mathrm{k}=1}^{\mathrm{n}} {\mathrm{k}}^2\text{, we get}\sum _{\mathrm{k}=1}^{\mathrm{n}} {\mathrm{k}}^2=\frac{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{6}$

Step-by-step solution

Step 1:

From the previous exercise 35, we have${\mathrm{a}}_{\mathrm{k}}={\mathrm{k}}^3$

Thus,$\sum _{k=1}^n \left(k^3-(k-1{)}^3\right)=n^3$

Expanding${(k-1)}^3$

$\sum _{k=1}^n \left(k^3-k^3+3k^2-3k+1\right)=n^3$

Step 2:

We shall cancel and reorganise, we get

$3\sum _{k=1}^n k^2=n^3+3\sum _{k=1}^n k-\sum _{k=1}^n 1$

Substituting the values from the exercise 37b, we get

$3\sum _{k=1}^n k^2=n^3+3\frac{n(n+1)}{2}-n$

Further putting together, the terms under a common denominator, we get

$\sum _{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}$