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)} = n^{2}+3n+1
S

_{n}= ∑t_{n}= ∑ (n^{2}+3n+1) = ∑n^{2 }+ 3∑n^{ }+∑1 = n(n+1)(2n+1)/6 + 3n(n+1)/2 + n
=> S

_{n}=n/6 (2n^{2}+1+3n+9n+9+6) = n(n^{2}+6n+8)/3
=> S

_{n}=n(n+2)(n+4)/3
Answer (4)

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