This is a geometric progression.
Let the 1st term be a and the common ratio r.
Then a=n1 and r=2n1÷n1=2n1×1n=21.
Let the nth term be Un. By the formula for the nth term of a geometric progression,
Un=arn−1
So the 16th term of the geometric progression is
U16=ar15=n1×(21)15=n1×327681=32768n1
Let the sum of the first n terms of the sequence be S16. Then
S16=1−ra(1−rn)
So the sum of the first 16 terms is
S16=1−ra(1−rn)=n11−21(1−(21)n)
=n1×2×6553665535=32768n65535
If n=156 then
Sn=51180865535
Comments