Question

1² + 2² + 3² + ... + n² =\[\frac{n(n + 1)(2n + 1)}{6}\] .

 

Answer 1

Let P(n) be the given statement.
Now,

\[P(n) = 1^2 + 2^2 + 3^2 + . . . + n^2 = \frac{n(n + 1)(2n + 1)}{6}\]

\[\text{ Step } 1: \]

\[P(1) = 1^2 = \frac{1(1 + 1)(2 + 1)}{6} = \frac{6}{6} = 1\]

\[\text{ Hence, P(1) is true}  . \]

\[\text{ Step } 2: \]

\[\text{ Let P(m) be true .}  \]

\[\text{ Then,}  \]

\[ 1^2 + 2^2 + . . . + m^2 = \frac{m(m + 1)(2m + 1)}{6}\]

\[\text{ We shall now prove that P(m + 1) is true}  . \]

\[i . e . , \]

\[ 1^2 + 2^2 + 3^2 + . . . + (m + 1 )^2 = \frac{(m + 1)(m + 2)(2m + 3)}{6}\]

\[ \text{ Now } , \]

\[P(m) = 1^2 + 2^2 + 3^2 + . . . + m^2 = \frac{m(m + 1)(2m + 1)}{6}\]

\[ \Rightarrow 1^2 + 2^2 + 3^2 + . . . + m^2 + (m + 1 )^2 = \frac{m(m + 1)(2m + 1)}{6} + (m + 1 )^2 \left[ \text{ Adding}  (m + 1 )^2 \text{ to both sides}  \right]\]

\[ \Rightarrow 1^2 + 2^2 + 3^2 + . . . + (m + 1 )^2 = \frac{m(m + 1)(2m + 1) + 6(m + 1 )^2}{6} = \frac{(m + 1)(2 m^2 + m + 6m + 6)}{6} = \frac{(m + 1)(m + 2)(2m + 3)}{6}\]

\[\text{ Hence, P(m + 1) is true } . \]

\[\text{ By the principle of mathematical induction, the given statement is true for all n } \in N .\]