JEE Main 2014 · 9 Apr · Shift 1 · Mathematics Q70

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MathematicsSingle correct+4 / -1Question 70 of 90

If the sum
$\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+\dots\text{ upto 20 terms}$
is equal to $\dfrac{k}{21}$ , then $k$ is equal to :

120

180

240

Answer and worked solution

Official answer: A

The $n$ th term is
$\frac{2n+1}{n(n+1)(2n+1)/6}=\frac6{n(n+1)}=6\left(\frac1n-\frac1{n+1}\right).$
The sum telescopes: $6\left(1-\frac1{21}\right)=\dfrac{120}{21}$ . So $k=120$ .

Topics:Special series sums

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JEE Main 2014 · 9 Apr · Shift 1, Mathematics question 70