The sum $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}$. What is the value of $21^2 + 22^2 + \dots + 40^2$?

Question

The sum $1^2 + 2^2 + 3^2 + 4^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$. What is the value of $21^2 + 22^2 + \cdots + 40^2$? Using induction, how can I solve this problem?

Answer 1

By your formula: $$1^2+2^2+\dots+40^2 - (1^2+\dots+20^2) = \frac{40\cdot 41 \cdot 81-20\cdot21\cdot 41}{6}.$$

Answer 2

You don't need induction. The sum from $21$ to $40$ is the sum from $1$ to $40$ less ???

Answer 3

HINT

We can consider

$$\sum_{n=1}^{40} n^2-\sum_{n=1}^{20} n^2$$

Answer 4

You've stated that: $$\sum_{x=1}^{n}{x^2}=\frac{1}{6}n(n+1)(2n+1)$$

$$\sum_{x=21}^{40}{x^2}=\sum_{x=1}^{40}{x^2}-\sum_{x=1}^{20}{x^2}$$ Simply use the formula to evaluate this.

Answer 5

You don't need induction to calculate that value. Value is simply $$(1^2+2^2+...40^2)-(1^2+2^2+...20^2)$$