Sum of the series 1-x-1-x-x2-1-x-x2-x3- upto n terms isA- None of these 1-1-x-n-x21-xn-1-x-C- 1-1-x-n-x31-xn-1-x-D- 1-1-x-n-x1-xn-1-x-

The correct option is B We have, (1+x)+(1+x+x^2)+(1+x+x^2+x^3)+......upto n terms =1 x^2/1 x+1 x^3/1 x+1 x^4/1 x+.... to n terms =1/1 x[(1+1+1+....n terms ...

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

Byju's Answer

Standard XI

Mathematics

Nth term of A.G.P

Sum of the se...

Question

Sum of the series $(1 + x) + (1 + x + x^{2}) + (1 + x + x^{2} + x^{3}) +$ .............. upto n terms is

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!

None of these

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

Open in App

Solution

The correct option is A
We~have,
$(1 + x) + (1 + x + x^{2}) + (1 + x + x^{2} + x^{3}) + . . . . . . u p t o n t e r m s = \frac{1 - x^{2}}{1 - x} + \frac{1 - x^{3}}{1 - x} + \frac{1 - x^{4}}{1 - x} + . . . . t o n t e r m s = \frac{1}{1 - x} [(1 + 1 + 1 + . . . . n t e r m s - (x^{2} + x^{3} + x^{4} + . . . t o n t e r m s)] = \frac{1}{1 - x} [n - \frac{x^{2} (1 - x^{n})}{1 - x}]$

Suggest Corrections

Similar questions

Q. If

(1 + x) (1 + x^{2}) (1 + x^{4}) . . . .

(1 + x^{128}) = (1 + x + x^{2} . . . . . x^{n})

then

n = ?

Given

| x | < 1

{lim}_{x \leftarrow 1} \frac{x + x^{2} + x^{3} + . . . . + x^{n} - n}{x - 1} =

Q. Given that

1 + x + x^{2} + \dots x^{n - 1} = \frac{1 - x^{n}}{1 - x}, x \neq 1

, Using this, find the sum of the series

1 + 2 x + 3 x^{2} + \dots (n - 1) x^{n - 2}

$The value of l i m x \to 1 \frac{x^{n} + x^{n - 1} + x^{n - 2} + . . . . . . . + x^{2} + x - n}{x - 1}$

Q. Sum of the series:

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

upto

n

terms

Join BYJU'S Learning Program

Sum of the series (1+x)+(1+x+x2)+(1+x+x2+x3)+.............. upto n terms is

The correct option is A We~have, (1+x)+(1+x+x2)+(1+x+x2+x3)+......upto n terms=1−x21−x+1−x31−x+1−x41−x+....to n terms=11−x[(1+1+1+....n terms−(x2+x3+x4+...to n terms)]=11−x[n−x2(1−xn)1−x]

Sum of the series $(1 + x) + (1 + x + x^{2}) + (1 + x + x^{2} + x^{3}) +$ .............. upto n terms is