If -k-1n fk-n2n-2- then the value of -k-1101fk is equal to

Let S_n=∑_k=1^n f(k)=n2(n+2) ⇒ S_n S_n 1=n2(n+2) n 12(n+1) ⇒ S_n S_n 1=1(n+1)(n+2) ⇒ f(n) =1(n+1)(n+2) ⇒1f(n)=(n+1)(n+2) ⇒1f(n) = n^2+3n+2 ∴∑_k=1^101f(k) =∑_k= ...

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Byju's Answer

Standard XIII

Mathematics

Vn Method

If ∑k=1n fk=n...

Question

If $n \sum k = 1 f (k) = \frac{n}{2 (n + 2)}$ , then the value of $10 \sum k = 1 \frac{1}{f (k)}$ is equal to

570

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480

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560

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550

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Solution

The correct option is A $570$
Let $S_{n} = n \sum k = 1 f (k) = \frac{n}{2 (n + 2)}$
$S_{n} - S_{n - 1} = \frac{n}{2 (n + 2)} - \frac{n - 1}{2 (n + 1)} \Rightarrow S_{n} - S_{n - 1} = \frac{1}{(n + 1) (n + 2)} \Rightarrow f (n) = \frac{1}{(n + 1) (n + 2)} \Rightarrow \frac{1}{f (n)} = (n + 1) (n + 2) \Rightarrow \frac{1}{f (n)} = n^{2} + 3 n + 2 10 \sum k = 1 \frac{1}{f (k)} = 10 \sum k = 1 (k^{2} + 3 k + 2) = 10 \sum k = 1 k^{2} + 3 10 \sum k = 1 k + 20 = \frac{10 (11) (21)}{6} + 3 \frac{10 (11)}{2} + 20 = 385 + 165 + 20 = 570$

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Vn Method

Standard XIII Mathematics

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If n∑k=1f(k)=n2(n+2), then the value of 10∑k=11f(k) is equal to

The correct option is A 570Let Sn=n∑k=1f(k)=n2(n+2) Sn−Sn−1=n2(n+2)−n−12(n+1)⇒Sn−Sn−1=1(n+1)(n+2)⇒f(n)=1(n+1)(n+2)⇒1f(n)=(n+1)(n+2)⇒1f(n)=n2+3n+210∑k=11f(k)=10∑k=1(k2+3k+2)=10∑k=1k2+310∑k=1k+20=10(11)(21)6+310(11)2+20=385+165+20=570

If $n \sum k = 1 f (k) = \frac{n}{2 (n + 2)}$ , then the value of $10 \sum k = 1 \frac{1}{f (k)}$ is equal to