Series-4 (NCERT)

Find sum to n terms of the following series:

5+11+19+29+41……….

(1) (n+1)(2n+1)(3n+2)/6 

(2) n(2n+1)(3n+2)/3  

(3) n(n+2)(2n+3)/3  

(4) n(n+2)(n+4)/3

(5) None of these

Answer follows here:

Solution:

This is a typical series problem.

We are trying to form a series in A.P.

Let S = 5+11+19+29+41………+t_n-1+t_n                     ------(1)

      S =        5+11+19+29+41………+t_n-1+t_n              ------(2)

Subtracting (1)-(2):

0 = 5+6+8+10+12+.....-t_n

=> t_n = 5+(6+8+10+12+....(n-1) terms)

Second part is A.P having (n-1) terms with initial term=a=6, common difference =d=2;

=> t_n = 5+{(n-1)/2 * [2(6)+(n-2)(2)]}

 => t_n = 5+{(n-1)(2n+8)/2} = 5+{(n-1)(n+4)} = n²+3n+1

S_n = ∑t_n = ∑ (n²+3n+1) = ∑n² + 3∑n +∑1 = n(n+1)(2n+1)/6  + 3n(n+1)/2 + n

=> S_n =n/6 (2n²+1+3n+9n+9+6) = n(n²+6n+8)/3

=> S_n =n(n+2)(n+4)/3

Answer (4)