A) 2925
B) 1469
C) 1728
D) 1456
Correct Answer: A
Solution :
Consider \[{{1}^{2}}+{{3}^{2}}+{{5}^{2}}+.....+{{25}^{2}}\] \[{{n}^{th}}\text{term}{{T}_{n}}={{(2n-1)}^{2}},n=1,.....13\] Now.\[{{S}_{n}}=\sum\limits_{n=1}^{13}{{{T}_{n}}}={{T}_{n}}=\sum\limits_{n=1}^{13}{{{(2n-1)}^{2}}}\] \[=\sum\limits_{n=1}^{13}{4{{n}^{2}}}+\sum\limits_{n=1}^{13}{1-\sum\limits_{n=1}^{13}{4n}}\] \[=4\sum\limits_{{}}^{{}}{{{n}^{2}}}+13-4\sum\limits_{{}}^{{}}{n}\] \[=4\left[ \frac{n(n+1)(2n+1)}{6} \right]+13-4\frac{n(n+1)}{2}\] Put\[n=13,\]we get \[{{S}_{n}}=26\times 14\times 9+13-26\times 14\] \[=3276+13-364\]\[=2925.\]
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