Question

Use Mathematical Induction to prove the following:
(i) 1
2 + 22 + 32 + … + n
2 =

Answer 1

Answer on question#50406

Use Mathematical Induction to prove the following:

$$1^{2} + 2^{2} + 3^{2} + \cdots + n^{2} = \frac{n(n + 1)(2n + 1)}{6}$$

Proof:

For $n = 1$, the statement reduces to $1^{2} = \frac{1 \cdot 2 \cdot 3}{6}$ and is obviously true. Assuming the statement is true for $n = k$:

$$1^{2} + 2^{2} + 3^{2} + \cdots + k^{2} = \frac{k(k + 1)(2k + 1)}{6},$$

we will prove that the statement must be true for $n = k + 1$:

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

The left-hand side of (2) can be written as

$$1^{2} + 2^{2} + 3^{2} + \cdots + k^{2} + (k + 1)^{2}.$$

In view of (1), this simplifies to:

$$\begin{aligned} 1^{2} + 2^{2} + 3^{2} + \cdots + k^{2} + (k + 1)^{2} &= \frac{(k + 1)(k + 2)(2k + 3)}{6} + (k + 1)^{2} \\ &= \frac{k(k + 1)(2k + 1) + 6(k + 1)^{2}}{6} \\ &= \frac{(k + 1)[k(2k + 1) + 6(k + 1)]}{6} = \frac{(k + 1)(2k2 + 7k + 6)}{6} \\ &= \frac{(k + 1)(k + 2)(2k + 3)}{6}. \end{aligned}$$

Thus the left-hand side of (2) is equal to the right-hand side of (2). This proves the inductive step. Therefore, by the principle of mathematical induction, the given statement is true for every positive integer $n$.

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