JEE Main 2015 · 11 Apr · Shift 1 · Mathematics Q69

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MathematicsSingle correct+4 / -1Question 69 of 90

If $\displaystyle\sum_{n=1}^5\frac1{n(n+1)(n+2)(n+3)}=\frac k3$ , then k is equal to :

\dfrac{55}{336}

\dfrac{17}{105}

\dfrac16

\dfrac{19}{112}

Answer and worked solution

Official answer: A

$\dfrac1{n(n+1)(n+2)(n+3)}=\dfrac13\left[\dfrac1{n(n+1)(n+2)}-\dfrac1{(n+1)(n+2)(n+3)}\right]$ , so the sum telescopes:
$\frac13\left(\frac16-\frac1{336}\right)=\frac13\cdot\frac{55}{336},\qquad k=\frac{55}{336}.$

Topics:Special series sums

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JEE Main 2015 · 11 Apr · Shift 1, Mathematics question 69