Question

1² + 3² + 5² + ... + (2n − 1)² = \[\frac{1}{3}n(4 n^2 - 1)\]

 

Answer 1

Let P(n) be the given statement.
Now,

\[P(n) = 1^2 + 3^2 + 5^2 + . . . + (2n - 1 )^2 = \frac{1}{3}n(4 n^2 - 1)\]

\[\text{ Step 1:}  \]

\[P(1) = 1^2 = 1 = \frac{1}{3} \times 1 \times (4 - 1)\]

\[\text{ Hence, P(1) is true }  . \]

\[\text{ Step 2: }  \]

\[\text{ Let P(m) be true } . \]

\[\text{ Then, } \]

\[ 1^2 + 3^2 + . . . + (2m - 1 )^2 = \frac{1}{3}m(4 m^2 - 1)\]

\[\text{ To prove: P(m + 1) is true whenever P(m) is true } . \]

\[\text{ That is, }  \]

\[ 1^2 + 3^2 = . . . + (2m + 1 )^2 = \frac{1}{3}(m + 1)\left\{ 4(m + 1 )^2 - 1 \right\}\]

\[\text{ We know that P(m) is true } . \]

\[\text{ Thus, we have: } \]

\[ 1^2 + 3^2 + . . . + (2m - 1 )^2 = \frac{1}{3}m(4 m^2 - 1)\]

\[ \Rightarrow 1^2 + 3^2 + . . . + (2m - 1 )^2 + (2m + 1 )^2 = \frac{1}{3}m(4 m^2 - 1) + (2m + 1 )^2 \left[ \text{ Adding } (2m + 1 )^2 \text{ to both sides } \right]\]

\[ \Rightarrow P(m + 1) = \frac{1}{3}\left( 4 m^3 - m + 12 m^2 + 12m + 3 \right)\]

\[ \Rightarrow P(m + 1) = \frac{1}{3}(4 m^3 - m + 8 m^2 + 4m + 4 m^2 + 8m + 3)\]

\[ = \frac{1}{3}(m + 1)(4 m^2 + 8m + 3) \]

\[ = \frac{1}{3}(m + 1)(4(m + 1 )^2 - 1)\]

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

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