The sum of the series 13-1-13-23-1-3-13-23-33-1-3-5- upto 16 terms is-

The correct option is C 446Here, T_n =∑ n^3/n/2[2×1+(n 1)2]=n^2(n+1)^2/4/n^2 ⇒ S_16=1/4∑_16^k=1(k+1)^2=446 ...

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

Byju's Answer

Standard XII

Mathematics

nth Term of A.P

The sum of th...

Question

The sum of the series $\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + . . . . . . .$ upto 16 terms is,

246

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

346

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

446

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

546

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

Open in App

Solution

The correct option is C 446
$H e r e, T_{n} = \frac{\sum n^{3}}{\frac{n}{2} [2 \times 1 + (n - 1) 2]} = \frac{\frac{n^{2} (n + 1)^{2}}{4}}{n^{2}} \Rightarrow S_{16} = \frac{1}{4} \sum_{16}^{k = 1} (k + 1)^{2} = 446$

Suggest Corrections

Similar questions

Q. Sum up to

16

terms of the series

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

Find the sum of the series $\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 12 t e r m s$

Find the sum of the following series upto n terms :

$\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + \dots \dots$

Q. Sum to n terms of the series

\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + . . . . .

Q. Sum up to

16

terms of the series

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

Join BYJU'S Learning Program

The sum of the series 131+13+231+3+13+23+331+3+5+....... upto 16 terms is,

The correct option is C 446Here, Tn=∑n3n2[2×1+(n−1)2]=n2(n+1)24n2⇒S16=14∑k=116(k+1)2=446

The sum of the series $\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + . . . . . . .$ upto 16 terms is,

The correct option is C 446
$H e r e, T_{n} = \frac{\sum n^{3}}{\frac{n}{2} [2 \times 1 + (n - 1) 2]} = \frac{\frac{n^{2} (n + 1)^{2}}{4}}{n^{2}} \Rightarrow S_{16} = \frac{1}{4} \sum_{16}^{k = 1} (k + 1)^{2} = 446$