Question

Find the value of `sum_("n" = 1)^oo 1/(2"n" - 1) (1/(9^("n" - 1)) + 1/(9^(2"n"- 1)))`

Answer 1

S_n = `1/1(1 + 1/9) + 1/3(1/9 + 1/9^2) + 1/5(1/9^2 + 1/9^5) + 1/7(1/9^3 + 1/9^7) + ...`

= `(1 + 1/3*1/9 + 1/5*1/9^2 + ...) + (1/9 + 1/3*1/9^3 + 1/5*1/9^5) ...`

= `(1 + 1/3* 1/3^2 + 1/5*1/3^4 + 1/7* 1/3^6 + ...) + [1/9 + 1/3(1/9)^3 + 1/5(1/9)^5 + ...]`

Multiplying and dividing the 1 summation by 3 we get

`3(1/3 + 1/3*1/3^3 + 1/5*1/3^5 + ...) + 1/9 + 1/3(1/9)^3 + ...`

= `[3(1/3) + (1/3)^3/3 + (1/3)^5/5 + ....] + [1/9 + (1/9)^3/3 + (1/9)^5/5 + ...]`

`{"we know" 1/2 log((1 + x)/(1 - x)) = x + x^3/3 + x^5/5 + ...}`

= `3 xx 1/2 log((1 + 1/3)/(1 - 1/3)) + 1/2 log((1 + 1/9)/(1 -  1/9))`

= `1/2{3log  (4/3)/(2/3)} + 1/2log  (10/9)/(8/9)`

= `1/2[3 log 2 + log  10/8]`

= `1/2[log2^3 + log  10/8]`

= `1/2[log 8* 10/8]`

= `1/2 log "e"^10`  .....[log a + log b = log ab]