JEE Main 2016 · 3 Apr · Shift 1 · Mathematics Q69

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MathematicsSingle correct+4 / -1Question 69 of 90

If the sum of the first ten terms of the series $\left(1\dfrac35\right)^2+\left(2\dfrac25\right)^2+\left(3\dfrac15\right)^2+4^2+\left(4\dfrac45\right)^2+\dots$ , is $\dfrac{16}5m$ , then m is equal to :

102

101

100

Answer and worked solution

Official answer: B

The terms are $\left(\frac{8}5\right)^2,\left(\frac{12}5\right)^2,\dots=\frac{16}{25}k^2$ for $k=2,\dots,11$ . Their sum is $\frac{16}{25}(506-1)=\frac{16}{25}\times505=\frac{16}5\times101$ , so $m=101$ .

Topics:Special series sums

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JEE Main 2016 · 3 Apr · Shift 1, Mathematics question 69