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Find the sum of the series `1+3+7+15+31+......` n terms. |
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Answer» The sequence of differences berween successive terms is `2,4,8,16,"…."`. Clearly, it is a GP with common ratio 2. So, let the nth term and sum of the series upto n terms of the series be `T_(n)` and `S_(n)`, respectively. Then, `S_(n)=1+3+7+15+31+"..."T_(n-1)+T_(n) "....(i)"` `S_(n)=1+3+7+15+"..."T_(n-1)+T_(n) "....(ii)"` Subtracting Eq.(ii) from Eq.(i), we get `0=1+2+4+8+16+"..."(T_(n)-T_(n-1))-T_(n)` ` implies T_(n)=1+2+4+8+16+"..." " upto n terms "` ` =(1*(2^(n)-1))/(2-1)` Hence, `T_(n)=(2^(n)-1)` `therefore` Sum of n terms `S_(n)=sumT_(n)=sum (2^(n)-1)=sum2^(n)-sum 1` `=(2+2^(2)+2^(3)+"...."+2^(n))-n` `=(2*(2-1))/(2-1)-n=2^(n+1)-2-n` |
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