Find the sum of n terms of a GP series whose m'th term is 2^m + 2m
mysticd:
it is not gp may be it is ap
No it is given in ml aggarwal. If it's ap then also help me to do the sum
Please help me to find the sum as mentioned in question
Answers
Answered by
110
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)
i hope this will useful to u
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)
i hope this will useful to u
Answered by
0
Step-by-step explanation:
answer is 2(2n-1)+n(n+1)
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