Question

prove that 1^2+2^2+3^2+-----n^2=n(n+1)(2n+1) for all positive integers

Answer 1

Answer on Question #60952 – Math – Discrete Mathematics

Question

Prove that $1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}$

Proof

We will use mathematical induction to prove this formula.

**Basis**: $n = 1$ (the least positive integer). The statement is

$$1^2 = \frac{1 \cdot 2 \cdot 3}{6}.$$

Obviously it's true.

**Inductive step**:

Assuming the statement is true for $n = k$:

$$1^2 + 2^2 + \ldots + k^2 = \frac{k(k+1)(2k+1)}{6}.$$

We must prove that the statement is true for $n = k + 1$:

$$1^2 + 2^2 + \ldots + k^2 + (k+1)^2 = \frac{(k+1)(k+2)(2k+3)}{6}.$$

**Consider**

$$\begin{aligned} (1^2 + 2^2 + \ldots + k^2) + (k+1)^2 &= \frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \\ &= \frac{k(k+1)(2k+1) + 6(k+1)^2}{6} = \frac{(k+1)(2k^2 + k + 6k + 6)}{6} = \\ &= \frac{(k+1)(2k^2 + 7k + 6)}{6} = \frac{(k+1)(k+2)(2k+3)}{6} \end{aligned}$$

Therefore, by the principle of mathematical induction, the given statement is true for every positive integer $n$.

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