-k-1n fk-n-2 n -2- then the value of -k-1101- f k is equal toA- 480- 560- 550D- 570

The correct option is D 570Let 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+ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Continuity of a Function

∑k=1n fk=n/2 ...

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

480

No worries! We‘ve got your back. Try BYJU‘S free classes today!

560

No worries! We‘ve got your back. Try BYJU‘S free classes today!

550

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. 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

Q. 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

The value of $\sum_{r = 1}^{n + 1} (\sum_{k = 1}^{n}^{k} C_{r - 1})$ where r, k, n $ϵ$ N is equal to

$\infty \sum n = 1 \frac{1}{(n + 1) (n + 2) (n + 3) . . . . (n + k)}$ is equal to

Q. If for

n \in N

\sum_{k = 0}^{2 n} {(- 1)}^{k} {({^{2 n} C}_{k})}^{2} = A

, then value of

\sum_{k = 0}^{2 n} {(- 1)}^{k} (k - 2 n) {({^{2 n} C}_{k})}^{2}

Join BYJU'S Learning Program

Explore more

Continuity of a Function

Standard XI Mathematics

Join BYJU'S Learning Program

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