1²+3²+5²++(2n-1)²=

Correct Answer is : n(4n2−1)3{n(4n²-1)}{3}3n(4n2−1)

Correct Answer is : n(4n2−1)3{n(4n²-1)}{3}3n(4n2−1)

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AP and TS EAMCET Engineering Model Paper 20

Question : 3 of 160

Marks: +1, -0

1^{2}+3^{2}+5^{2}+\dots+(2n-1)^{2}=

Solution:

1^{2}+3^{2}+5^{2}+\dots+(2n-1)^{2}

= Sum of squares of n odd natural numbers

= \left[ \{1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+\dots \} \right]

+ (2n-1)^{2}+(2n)^{2} \} - \{2^{2}+4^{2}+6^{2}+\dots

\left[ + (2n)^{2} \} \right]

\frac{2n(2n+1)[2(2n)+1]}{6}

-2^{2} \left[ \frac{n(n+1)(2n+1)}{6} \right]

= \frac{n(2n+1)(4n+1)}{3}

- \frac{2[n(n+1)(2n+1)]}{3}

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

= \frac{n(2n+1)(2n-1)}{3}

= \frac{n(4n^{2}-1)}{3}

\text{or} \left( \sum(2n)^{2} - 2^{2} \sum(n)^{2} = \frac{n(4n^{2}-1)}{3} \right)

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