JEE Main 2015 · 4 Apr · Shift 1 · Mathematics Q69

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MathematicsSingle correct+4 / -1Question 69 of 90

The sum of first 9 terms of the series
$\frac{1^3}1+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots$
is :

142

192

Answer and worked solution

Official answer: B

The $n$ th term is $\dfrac{\left(\frac{n(n+1)}2\right)^2}{n^2}=\dfrac{(n+1)^2}4$ .
$\sum_{n=1}^9\frac{(n+1)^2}4=\frac{(2^2+\dots+10^2)}4=\frac{385-1}4=96.$

Topics:Special series sums

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JEE Main 2015 · 4 Apr · Shift 1, Mathematics question 69