Question
Mathematics Question on Arithmetic Progression
The sum of 1st n terms of the series 112+1+212+22+1+2+312+22+32+...
A
\ce3n+2
B
\ce3n(n+2)
C
\ce3n(n−2)
D
\ce6n(n−2)
Answer
\ce3n(n+2)
Explanation
Solution
We have, 112+1+212+22+1+2+312+22+32+… ∴an=1+2+3+…+n12+22+32+…+n2 ⇒an=2n(n+1)6n(n+1)(2n+1) ⇒[∵Σn2=6n(n+1)(2n+1)andΣn=2n(n+1)∴=32n+1] an=32n+1 Now, Sn=Σan=Σ32n+1=31[∑2n+Σ1] =31[22n(n+1)+n][∵Σ1=n] =31[n2+n+n] =31(n2+2n)=3n(n+2)