12 + 32 + 52 + ... + (2n − 1)2 = 1 3 N ( 4 N 2 − 1 ) - Mathematics

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

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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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 12 Mathematical Induction
Exercise 12.2 | Q 16 | Page 27