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

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

 
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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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APPEARS IN

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