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