Question

1+5+13+29+61+.............n terms then find sum to n terms

Answer 1

Answer:

Let S=1+5+13+29+.....+U n-1 +U n

again S=1+5+13+.....+Un-2 +Un-1 +Un

these are two equations

hence equation (1)-(2)

=»0=(1+4+8+16+....nth term) - Un

=»Un = 1+4+8+16+....nth term

= 1+[4+8+16+.....(n-1)th term]

= 1+ 4(2^n-1 -1)/2-1

= 1+4.2^n-1 -4

= 2^n+1 -3

the sum of the serise

=(2^2+2^3+2^4+....+2^n+1) - (3+3+3+....nth term)

=4(2^n -1)/2-1 -3n

=4(2^n -1) -3n

= 2^n+2 -3n -4