Manipal MET2015MathematicsSequences and Series
The sum to n terms of the series 3 1^2 + 5 1^2+2^2 + 7 1^2+2^2+3^2 + , is
Options
- A6 n n+1
- B9 n n+1
- C12 n n+1
- D3 n n+1
Correct answer
A. 6 n n+1
Step-by-step solution
Let T_r be the r th term of the given series. Then, T_r= 2 r+1 1^2+2^2+ +r^2 = 2 r+1 r(r+1)(2 r+1) 6 = 6 r(r+1) T_r=6 ( 1 r - 1 r+1 ) So, the required sum is given by _ r=1 ^n T_r=6 _ r=1 ^n ( 1 r - 1 r+1 ) aligned & =6 [ 1 1 - 1 2 + 1 2 - 1 3 + + 1 n - 1 n+1 ] & =6 [1- 1 n+1 ]= 6 n n+1 aligned