Question

Find the sum of n teems of a series whose nth term is 2m+2m

Answer 1

Given

tm = 2^m + 2m

∴tn = 2^n +2n

sum of n terms = ∑tn

=∑(2^n +2n)

=∑2^n + ∑2n

make it two parts

i) ∑2^n = [2+2²+2³+..........+2^n]

series is GP

here first term =a =2

common ratio =r = t2/t1 = 2²/2 = 2 >1

sum of n terms in GP = a(r^n -1)/(r-1)

= [2(2^n -1]/(2-1)

=2(2^n -1)------(1)

ii) sum of n terms =∑ 2n = 2∑n=2[1+2+3+....+n]

=2n(n+1)/2 since sum of n natural numbers = n(n+1)/2

=n(n+1)------(2)

therefore

∑tn sum of n terms in given series = (1) +(2)

= 2(2^n - 1) +n(n+1)

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