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Question
12 + 32 + 52 + ... + (2n − 1)2 = \[\frac{1}{3}n(4 n^2 - 1)\]
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Solution
Let P(n) be the given statement.
Now,
\[P(n) = 1^2 + 3^2 + 5^2 + . . . + (2n - 1 )^2 = \frac{1}{3}n(4 n^2 - 1)\]
\[\text{ Step 1:} \]
\[P(1) = 1^2 = 1 = \frac{1}{3} \times 1 \times (4 - 1)\]
\[\text{ Hence, P(1) is true } . \]
\[\text{ Step 2: } \]
\[\text{ Let P(m) be true } . \]
\[\text{ Then, } \]
\[ 1^2 + 3^2 + . . . + (2m - 1 )^2 = \frac{1}{3}m(4 m^2 - 1)\]
\[\text{ To prove: P(m + 1) is true whenever P(m) is true } . \]
\[\text{ That is, } \]
\[ 1^2 + 3^2 = . . . + (2m + 1 )^2 = \frac{1}{3}(m + 1)\left\{ 4(m + 1 )^2 - 1 \right\}\]
\[\text{ We know that P(m) is true } . \]
\[\text{ Thus, we have: } \]
\[ 1^2 + 3^2 + . . . + (2m - 1 )^2 = \frac{1}{3}m(4 m^2 - 1)\]
\[ \Rightarrow 1^2 + 3^2 + . . . + (2m - 1 )^2 + (2m + 1 )^2 = \frac{1}{3}m(4 m^2 - 1) + (2m + 1 )^2 \left[ \text{ Adding } (2m + 1 )^2 \text{ to both sides } \right]\]
\[ \Rightarrow P(m + 1) = \frac{1}{3}\left( 4 m^3 - m + 12 m^2 + 12m + 3 \right)\]
\[ \Rightarrow P(m + 1) = \frac{1}{3}(4 m^3 - m + 8 m^2 + 4m + 4 m^2 + 8m + 3)\]
\[ = \frac{1}{3}(m + 1)(4 m^2 + 8m + 3) \]
\[ = \frac{1}{3}(m + 1)(4(m + 1 )^2 - 1)\]
\[\text{ Thus, P(m + 1) is true } . \]
\[\text{ By the principle of mathematical induction, P(n) is true for all n } \in N . \]
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