Evaluate
3+5+9+17+33+.....to n terms
Hi,
This question can be done easily with the help of a method called 'Method of Differences'
Let S be the sum of the given series.
S=3+5+9+17+33... T(n) -->(1)
S= 3+5+ 9+ 17... T(n-1)+ T(n) -->(2)
NOTE: 'n', 'n-1' aren't multiplied here. They are used just to denote the (n-1)th and nth term.
Subtracting 2 from 1, we get
0= 3+ 2+4+8+16.... [T(n) -T(n-1)] -T(n)
=> T(n)= 3+ 2+4+8+16... [T(n) -T(n-1)
T(n)= 3+ { 2+4+8+16...}
The right most terms inside braces are in GP.
We can apply the formula for sum of a GP here
=> T(n)= 3+ { 2x [ 2^(n-1)-1]/(2-1) }
NOTE: here, r=2, a=2 and no. of terms= n-1
=> T(n)= 3+ 2^n-2
T(n)= 2^n+1
This is the nth term of the given series.
Sum till n terms: { 2, 4,8,16...} +n
= 2x[2^n-1]/(2-1) +nn
SUM = 2^(n+1) - 2 +n
Regards
This question can be done easily with the help of a method called 'Method of Differences'
Let S be the sum of the given series.
S=3+5+9+17+33... T(n) -->(1)
S= 3+5+ 9+ 17... T(n-1)+ T(n) -->(2)
NOTE: 'n', 'n-1' aren't multiplied here. They are used just to denote the (n-1)th and nth term.
Subtracting 2 from 1, we get
0= 3+ 2+4+8+16.... [T(n) -T(n-1)] -T(n)
=> T(n)= 3+ 2+4+8+16... [T(n) -T(n-1)
T(n)= 3+ { 2+4+8+16...}
The right most terms inside braces are in GP.
We can apply the formula for sum of a GP here
=> T(n)= 3+ { 2x [ 2^(n-1)-1]/(2-1) }
NOTE: here, r=2, a=2 and no. of terms= n-1
=> T(n)= 3+ 2^n-2
T(n)= 2^n+1
This is the nth term of the given series.
Sum till n terms: { 2, 4,8,16...} +n
= 2x[2^n-1]/(2-1) +nn
SUM = 2^(n+1) - 2 +n
Regards