InterviewSolution
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InterviewSolution
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Find the sum of n terms of the series 1+4+10+20+35+… |
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Answer» `S_(n)=1+4+10+20+35+...+T_(n-1)+T_(n)` `underline(S_(n)=""1+4+10+20+...+T_(n-1)+T_(n))` On SUBTRACTING `0=1+3+6+10+15+...+t_(n)-T_(n)` `rArr" "T_(n)=1+3+6+10+15+...+t_(n)` `underline(T_(n)=""1+3+6+10+...+t_(n-1)+t_(n))` On subtracting `0=[1+2+3+4+5+..."nth term"]-t_(n)` `rArr" "t_(n)=Sigman=(n(+1))/(2)` `=(1)/(2)(n^(2)+n)` `rArr" "T_(n)=(1)/(2)(Sigman^(2)+Sigman)` `=(1)/(2)[(1)/(6)n(n+1)(2n+1)+(1)/(2)n(n+1)]` `=(1)/(2)n(n+1)[(2n+1+3)/(6)]` `=(1)/(6)n(n+1)(n+2)` `rArr" "T_(n)=(1)/(6)[n^(2)+3n^(2)+2n]` `rArr" "S_(n)=(1)/(6)[Sigman^(2)+3Sigman^(2)+2Sigman]` `=(1)/(6)[(1)/(4)n^(2)(n+1)^(2)+(3)/(6)n(n+1)(2n+1)+(2)/(2)n(n+1)]` `=(1)/(6)n(n+1)[(1)/(4)n(n+1)+(1)/(2)(2n+1)+1]` `=(1)/(6)n(n+1)[(n(n+1)+2(2n+1)+4)/(4)]` `=(1)/(24)n(n+1)(n^(2)+n+4n+2+4)` `=(1)/(24)n(n+1)(n^(2)+5n+6)` `=(1)/(24)n(n+1)(n+2)(n+3)` |
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