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Find the sum up to 7^th term of series 2+3+5+8+12+………………….(a) 70(b) 490(c) 340(d) 420The question was asked in an interview for job.This interesting question is from Sum to n Terms of Special Series in portion Sequences and Series of Mathematics – Class 11

Answer»

The correct choice is (a) 70

Easiest EXPLANATION: Sn = 2+3+5+8+12+……………………………+ an

Sn = 2+3+5+8+12+ ……. + an-1 + an

Subtracting we get, 0 = 2+1+2+3+4+………………………….. – an

=>an = 2+1+2+3+4+…………….+(N-1) = 2+(n-1)n/2 = (1/2) (n^2-n+4)

n^th TERM is (1/2) (n^2-n+4)

So, AK = (1/2) (k^2-k+4)

Taking summation from k=1 to k=n on both sides, we get

\(\sum_{i=0}^na_k = (1/2)\sum_{i=0}^nk^2 – (1/2)\sum_{i=0}^nk + 2n\) = n(n+1) (2n+1)/(2*6) – n(n+1)/4 + 2n

Here, n=7. So, \(\sum_{i=0}^na_k\) = (7*8*15)/12 – (7*8)/4 + 2*7 = 70.


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