InterviewSolution
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InterviewSolution
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Find the sum upto n terms of the series `1*4*7*10*13*16+"...."`. |
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Answer» Let `T_(n)` be the nth term of the given series. ` therefore T_(n)=(" nth term of " 1,4,7, "..."(" nth term of " 4,7,10,"......")` `(" nth term of " 7,10,13, "...")(" nth term of " 10,13,16,"......")` `T_(n)=(3n-2)(3n+1)(3n+4)(3n+7)" " ".........(i)"` ` therefore V_(n)=(3n-2)(3n+1)(3n+4)(3n+7)(3n+10)` ` V_(n-1)=(3n-5)(3n-2)(3n+1)(3n+4)(3n+7)` `implies V_(n)=(3n+10)T_(n) " " [" from Eq. (i) "]` and ` V_(n-1)=(3n-5)T_(n)` ` therefore V_(n)-V_(n-1)=15T_(n)` `therefore T_(n)=(1)/(15)(V_(n)-V_(n-1))` `therefore S_(n)=sumT_(n)=sum_(n=1)^(n)(1)/(15)(V_(n)-V_(n-1))` `=(1)/(15)(V_(n)-V_(0)) " " [" from important Theorem 1 of " sum]` `=(1)/(15){(3n-2)(3n+1)(3n+4)(3n+7)(3n+10)-(-2)(1)(4)(7)(10)}` `=(1)/(15){(3n-2)(3n+1)(3n+4)(3n+7)(3n+10)+560}` Shortcut Method `S_(n)=(1)/(" last factor of III term - first factor of I term ")` ( Taking one extra factor in `T_(n)` in last - Taking one exra factor in I term in start ) `=(1)/((16-1)){(3n-2)(3n+1)(3n+4)(3n+7)(3n+10)-(-2)*1*4*7*10}` `=(1)/(15){(3n-2)(3n+1)(3n+4)(3n+7)(3n+10)+560}` |
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