## Monday, 31 October 2011

### Series-3 (NCERT)

Find sum to n terms of the following series:
12 + (12+22) + (12+22+32) +……….
(1) n(n+1)(n+2)2/12
(2) n2(n+1)(n+2)/12
(3) n(n+1)2(n+2)/12
(4) n(n+1)(2n+1)/6
(5) n2(n+1)2/4
Solution:
The formulae to be used here are:
Sum of ‘n’ consecutive integers starting from 1 = 1+2+3+…..+n =  ∑n = n(n+1)/2
Sum of squares of ‘n’ consecutive integers starting from 1
= 12+22+32+…..+n2 = ∑n2 = n(n+1)(2n+1)/6
Sum of cubes of ‘n’ consecutive integers starting from 1
= 13+23+33+…..+n3 = ∑n3 = n2(n+1)2/4
Given series is:
12 + (12+22) + (12+22+32) +……….
nth term of the series = tn = 12+22+32+…..+n2  = ∑n2 = n(n+1)(2n+1)/6
Sum to n terms = ∑tn = ∑n(n+1)(2n+1)/6
= 1/6 ∑(n2+n)(2n+1)
= 1/6 ∑(2n3+ n2+2n2+n)
= 1/6 ∑(2n3+3n2+n)
= 1/3 ∑n3 +  1/2 ∑n2 +  1/6 ∑n
= 1/3 (n2(n+1)2/4) + ½ (n(n+1)(2n+1)/6) + 1/6 (n(n+1)/2)
= n(n+1)/12 {n(n+1)+ (2n+1) + 1}
= n(n+1)/12 {n2+n+2n+2}
= n(n+1)/12 {n(n+1)+2(n+1)}
= n(n+1)/12 {(n+1)(n+2)}
= n(n+1)2(n+2)/12

Short cut method:

As “None of these” option is not given, we can safely check the answer options by substituting value of ‘n’:
For n =1, LHS = 12 = 1
For RHS answer options can be checked one by one:
Option(1) ----   1(1+1)(1+2)2/12 = 1*2*9/12 ≠ 1 => this option is wrong
Option(2) ----   12(1+1)(1+2)/12 = 1*2*3/12 ≠ 1 => this option is wrong
Option(3) ----   1(1+1)2(1+2)/12 = 1*4*3/12 = 1 => this option is correct
Option(4) ----   1(1+1)(2(1)+1)/6 = 1*2*3/6 = 1 => this option is correct
Option(5) ----   12(1+1)2/4 = 1*4/4 = 1 => this option is correct
Left out options are (3),(4) and (5).
Now we check for n=2:
LHS = 12 + (12+22)= 1+5 = 6
Option(3) ----   2(2+1)2(2+2)/12 = 2*9*4/12 = 6 => this option is correct
Option(4) ----   2(2+1)(2(2)+1)/6 = 2*3*5/6 ≠ 6 => this option is wrong
Option(5) ----   22(2+1)2/4 = 4*9/4 ≠ 6 => this option is wrong
The only left out option is (3)