Monday, 31 October 2011

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………+tn-1+tn                     ------(1)
      S =        5+11+19+29+41………+tn-1+tn              ------(2)
Subtracting (1)-(2):
0 = 5+6+8+10+12+.....-tn
=> tn = 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;
=> tn = 5+{(n-1)/2 * [2(6)+(n-2)(2)]}
 => tn = 5+{(n-1)(2n+8)/2} = 5+{(n-1)(n+4)} = n2+3n+1
Sn = ∑tn = ∑ (n2+3n+1) = ∑n2 + 3∑n +∑1 = n(n+1)(2n+1)/6  + 3n(n+1)/2 + n
=> Sn =n/6 (2n2+1+3n+9n+9+6) = n(n2+6n+8)/3
=> Sn =n(n+2)(n+4)/3
Answer (4)

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