Sequences & SerieseasyFree

Sequences & Series: Equal

JEE Maths question with a full step-by-step solution.

Question

a_k=\frac{1}{k(k+1)}

for

k=1,2,\ldots,n

, then

\left(\sum_{k=1}^{n}a_k\right)^2

is equal to

\frac{n}{n+1}

\frac{n^2}{(n+1)^2}

correct

\frac{n^4}{(n+1)^4}

\frac{n^6}{(n+1)^6}

Solution

Step 1:

a_k=\dfrac1k-\dfrac{1}{k+1}

, so the sum telescopes:

\sum_{k=1}^{n}a_k=1-\frac{1}{n+1}=\frac{n}{n+1}.

Step 2: Square:

\left(\sum_{k=1}^{n}a_k\right)^2=\frac{n^2}{(n+1)^2}.

Correct answer: (2)

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