wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let Sn=4nk=1(1)k(k+1)2. Then Sn can take value(s).

A
1056
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1088
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1120
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1332
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
A 1056
D 1332
4nk=1(1)k(k+1)2k2=1222+32+425262+72+82.......+(4n)2=(12+3252+72....+(4n1)2)+(22+4262+82....+(4n)2)=2(4+12+20+....(8n4))ntermsinA.P.+2(6+14+22+....(8n2))ntermsinA.P.=n(8n)+n(8n+4).....(usesumofA.P.formula)=n(16n+4)
.
Substituting n=8 gives the sum as 1056. Hence, option A is possible.
Similarly, substituting n=9 gives the sum as 1332. Hence, option D is also possible.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Special Integrals - 1
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon