Sum to n terms of the series 1-1-x1-2 x-1-1-2 x1-3 x-1-1-3 x1-4 x- is

The correct option is B If t_r denotes the nth term of the series, then nbsp; xt_r = nbsp; x/(1+rx)(1+(r+1)x) = nbsp; 1/1+rx nbsp; 1/1+(r+1)x ⇒ x ∑_ ...

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

Byju's Answer

Standard XII

Mathematics

Arithmetic Progression

Sum to n term...

Question

Sum to n terms of the series $\frac{1}{(1 + x) (1 + 2 x)}$ + $\frac{1}{(1 + 2 x) (1 + 3 x)}$ + $\frac{1}{(1 + 3 x) (1 + 4 x)}$ + ..............is

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

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

Open in App

Solution

The correct option is B

If $t_{r}$ denotes the nth term of the series, then

x $t_{r}$ = $\frac{x}{(1 + r x) (1 + (r + 1) x)}$ = $\frac{1}{1 + r x}$ - $\frac{1}{1 + (r + 1) x}$

⇒ x $n \sum r - 1 t_{r}$ = $n \sum r - 1$ [ $\frac{1}{1 + r x}$ - $\frac{1}{1 + (r + 1) x}$ ] = $\frac{1}{1 + x}$ - $\frac{1}{1 + (n + 1) x}$ = $\frac{n x}{(1 + x) (1 + (n + 1) x)}$

⇒ $n \sum r - 1 t_{r}$ = $\frac{n}{(1 + x) [1 + (n + 1) x]}$

Suggest Corrections

Similar questions

Q. Sum to

n

terms the series

\frac{1}{(1 + x) (1 + 2 x)} + \frac{1}{(1 + 2 x) (1 + 3 x)} + \frac{1}{(1 + 3 x) (1 + 4 x)} + . . . . .

Q. Sum to

n

terms of the series

\frac{1}{(1 + x) (1 + 2 x)} + \frac{1}{(1 + 2 x) (1 + 3 x)} + \frac{1}{(1 + 3 x) (1 + 4 x)} + . . .

Q. The sum to

n

terms of the series

\frac{1}{(1 + x) (1 + 3 x)} + \frac{1}{(1 + 3 x) (1 + 5 x)} + \frac{1}{(1 + 5 x) (1 + 7 x)} + \dots \dots

, where

n > 5, x \geq 2

Q. Sum to

n

terms the following series:

1 + 2 x + 3 x^{2} + 4 x^{3} + . . ., | x | < 1

Q. Find the sum to

n

terms of the series

1 + 2 x + 3 x^{2} + 4 x^{3} + . . .

Join BYJU'S Learning Program

Sum to n terms of the series 1(1+x)(1+2x) + 1(1+2x)(1+3x) + 1(1+3x)(1+4x) + ..............is

The correct option is B If tr denotes the nth term of the series, then xtr = x(1+rx)(1+(r+1)x) = 11+rx - 11+(r+1)x ⇒ x n∑r−1tr = n∑r−1[ 11+rx - 11+(r+1)x ] = 11+x - 11+(n+1)x = nx(1+x)(1+(n+1)x) ⇒ n∑r−1tr = n(1+x)[1+(n+1)x]

Sum to n terms of the series $\frac{1}{(1 + x) (1 + 2 x)}$ + $\frac{1}{(1 + 2 x) (1 + 3 x)}$ + $\frac{1}{(1 + 3 x) (1 + 4 x)}$ + ..............is