Question

Find the sum to n terms of the following series :
(i) (1) + (1+3)+(1+3+32) + .
(ii) 1 + 3 + 7 + 15 +31 + ...
(iii) 3 + 7 + 14 + 27 +52 + ...​

Answer 1

Solution:-

ii) 1+3+7+15+32+......

we have to find the sum of nth term of the series

Let S1=1+3+7+15+33+.....

shift one digit forward and then subtract the given terms

S1=1+3+7+15+31+....+(n-1)th+nth term.......i)

S1=. 1+3+7+ 15+....+(n-2)th+(n-1)th +(n)th term....ii)

subtracting equation ii) from I) we get

0=1+(2+4+8+16+........+(n-1)term) -nth term

nth term=1+(2+4+8+16+......+(n-1)term)

Tn=1+2×(2^(n-1) -1)/1

Tn=1+2^n-2

Tn=2^n-1

this will be nth term of the series

now,

sum upto n terms=2×(2^n-1)/(2-1) -n

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