The sum of the series 1+(1+2)+(1+2+3)+...........upto n terms,will be

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

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

Open in App

Solution

The correct option is D

Here $T_{n}$ = $\frac{n (n + 1)}{2}$

Therefore $S_{n}$ = $\frac{1}{2}$ [ $\sum n^{2} + \sum n$ ] = $\frac{n (n + 1) (n + 2)}{6}$

Suggest Corrections

Similar questions

1^{2} + 3^{2} + 5^{2} +

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

\frac{3}{1^{2}} + \frac{5}{1^{2} + 2^{2}} + \frac{7}{1^{2} + 2^{2} + 3^{2}} + . . . . .

n

1 + \frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + \frac{1 + 2 + 3 + 4}{4} +

^{'} n^{'}

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

Join BYJU'S Learning Program