The sum to n terms of the services 3-12-5-12-22-7-12-22-32- is -

The correct option is C 6n/n+1 General term =(2n+1)6(n)(n+1)(2n+1)=6n(n+1)=6[1n 1n+1] .Sum =6[1 12+12 13............ 1n+1] =6[1 1n+1]=6nx+1 ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

The sum to ...

Question

The sum to $n$ terms of the services $\frac{3}{1^{2}} + \frac{5}{1^{2} + 2^{2}} + \frac{7}{1^{2} + 2^{2} + 3^{2}} + . . . . . . .$ is -

\frac{3 n}{n + 1}

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\frac{6 n}{n + 1}

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\frac{9 n}{n + 1}

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\frac{12 n}{n + 1}

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Solution

The correct option is C $\frac{6 n}{n + 1}$
General term $= \frac{(2 n + 1) 6}{(n) (n + 1) (2 n + 1)} = \frac{6}{n (n + 1)} = 6 [\frac{1}{n} - \frac{1}{n + 1}]$ .
Sum $= 6 [1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} . . . . . . . . . . . . - \frac{1}{n + 1}]$
$= 6 [1 - \frac{1}{n + 1}] = \frac{6 n}{x + 1}$
$B$ is correct

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Similar questions

Q. Show that

\frac{1 \times 2^{2} + 2 \times 3^{2} + . . . + n \times {(n + 1)}^{2}}{1^{2} \times 2 + 2^{2} \times 3 + . . . + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}

Show that

$\frac{1 \times 2^{2} + 2 \times 3^{2} + \dots \dots + n \times (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + \dots \dots + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}$

Find the sum of $1. n^{2} + 2 (n - 1)^{2} + 3 (n - 2)^{2} + . . . . . . n {.1}^{2}$

Find $(3^{3} - 2^{3}) + (5^{3} - 4^{3}) + (7^{3} - 6^{3}) + . . . .$ to 10 terms.

Show that $\frac{1 \times 2^{2} + 2 \times 3^{2} + . . . . + n \times (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + . . . . + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}$

Q. Show that

1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . . . . . . . .

upto

n

terms

= \frac{n (n + 1)^{2} (n + 2)}{12}, \forall n \in N

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The sum to n terms of the services 312+512+22+712+22+32+....... is -

The correct option is C 6nn+1General term =(2n+1)6(n)(n+1)(2n+1)=6n(n+1)=6[1n−1n+1].Sum =6[1−12+12−13............−1n+1]=6[1−1n+1]=6nx+1B is correct

The sum to $n$ terms of the services $\frac{3}{1^{2}} + \frac{5}{1^{2} + 2^{2}} + \frac{7}{1^{2} + 2^{2} + 3^{2}} + . . . . . . .$ is -