Question

Compute the sum of first n terms of 1 + (1 + 4) + (1 + 4 + 4²) + (1 + 4 + 4² + 4³) + ...

Answer 1

T₁ = 1

T₂ = 1 + 4

T₃ = 1 + 4 + 4²

T_n = 1 + 4 + 4² + ... n terms which is a G.P.

∴ T_n = `(1(4^"n" - 1))/(4 - 1)`

= `(4^"n" - 1)/3`

So, S_n = `sum"t"_"n"`

= `(sum4^"n" - 1)/(sum3)`

= `(sum4^"n" - sum1)/(sum3)`

`sum4^"n" = 4^1 + 4^2 + ....... "n terms"`

= `(4(4^"n" - 1))/(4 - 1)`

= `(4(4^"n" - 1))/3`

So `(sum4^"n" - sum1)/(sum3)`

= `((4(4^"n" - 1)- "n")/3)/3`

= `(4(4^"n" - 1) - 3"n")/9`

= `4/9(4^"n" - 1) - "n"/3`