The sum of the series 3-12 - 5-12 - 22 - 7-12 - 22 - 32 - - upto n terms

The correct option is B 6n/n + 1 General term of the given series is, T_n=2n+1/∑_k=1^nk^2=2n+1/1/6n(n+1)(2n+1)=6/n(n+1) Thus , required summation is ...

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

Byju's Answer

Standard XII

Mathematics

Arithmetic Progression

The sum of th...

Question

The sum of the series
$\frac{3}{1^{2}} + \frac{5}{1^{2} + 2^{2}} + \frac{7}{1^{2} + 2^{2} + 3^{2}} + . . . . .$ upto n terms

\frac{2 n}{n + 1}

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

\frac{3 n}{n + 1}

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

\frac{3 n}{2 (n + 1)}

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

\frac{6 n}{n + 1}

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

Open in App

Solution

The correct option is B $\frac{6 n}{n + 1}$
General term of the given series is,
$T_{n} = \frac{2 n + 1}{\sum_{k = 1}^{n} k^{2}} = \frac{2 n + 1}{1 / 6 n (n + 1) (2 n + 1)} = \frac{6}{n (n + 1)}$
Thus , required summation is
$= n \sum k = 1 \frac{6}{k (k + 1)} = 6 n \sum k = 1 (\frac{1}{k} - \frac{1}{k + 1})$
$= 6 [(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) . . . . . . . . + (\frac{1}{n} - \frac{1}{n + 1})]$
$= 6 (1 - \frac{1}{n + 1}) = \frac{6 n}{n + 1}$
Hence option 'D' is correct choice.

Suggest Corrections

Similar questions

The sum of the following series, $\frac{1}{3}^{n} c_{1} + \frac{(1^{2} + 2^{2})}{5}^{n} c_{2} + \frac{(1^{2} + 2^{2} + 3^{2})}{7}^{n} c_{3}$ + ...... upto n terms is

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

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

Q. The sum of series

\frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + . . . .

upto n terms is

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

Join BYJU'S Learning Program

The sum of the series312+512+22+712+22+32+..... upto n terms

The correct option is B 6nn+1General term of the given series is,Tn=2n+1∑nk=1k2=2n+11/6n(n+1)(2n+1)=6n(n+1)Thus , required summation is =n∑k=16k(k+1)=6n∑k=1(1k−1k+1)=6[(11−12)+(12−13)+(13−14)........+(1n−1n+1)]=6(1−1n+1)=6nn+1Hence option 'D' is correct choice.

The sum of the series
$\frac{3}{1^{2}} + \frac{5}{1^{2} + 2^{2}} + \frac{7}{1^{2} + 2^{2} + 3^{2}} + . . . . .$ upto n terms