Find the sum upto n terms of the series `1*4*7*10*13*16+"...."`.

Explanation

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}`