CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question


lf Sn denotes the sum of first 'n' natural numbers then S1+S2x+S3x2+...+Snxn1+=

A
(1x)1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(1x)2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(1x)3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
(1x)4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B (1x)3

The given question can be written as.
1+(1+2)x+(1+2+3)x2+(1+2+3+4)x3...
=1+3x+6x2+12x3...
(1x)n=1n(x)+n(n1)x22!...
=1+nx+n(n+1)x22!...
Assuming x=1
Comparing the coefficients with that given in the question, we get.
nx=3 ...(i)
n(n+1)x22!=6 ...(ii)
Substituting the value of nx in equation (i), we get
3(nx+x)=12
3+x=4
x=1
Since nx=3
n=3
If we do not substitute the value of x and only substitute the value of n in the original equation, we get
1+3x+6x2+12x3...=(1x)n
=(1x)3


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Visualising the Terms
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon