Question

Sum to n terms of the following series: 1.3.5+2.4.6+3.6.9 +...

Answer 1

First note that
\[\large \begin{aligned} &\phantom{=} 1\cdot 3\cdot 5+2\cdot 4\cdot 6+3\cdot 6\cdot 9+\ldots\\ &=1\cdot 3\cdot 5+2^3(1\cdot 2\cdot 3)+3^3(1\cdot 2\cdot 3)+\ldots\end{aligned}\]This suggests that the rest of the terms will be of the form \( \large k^3(1\cdot 2\cdot 3)=6k^3\). However, that first term throws off everything. We can fix that though by noting that \(1\cdot3\cdot5=15=9+6=9+(1\cdot 2\cdot 3)\). Therefore, we see that \[\large \begin{aligned} &\phantom{=} 1\cdot3\cdot5+2\cdot4\cdot6+3\cdot6\cdot9+\ldots+n(1\cdot2\cdot 3)\\ &=1\cdot3\cdot5+2^3\cdot 6+3^3\cdot 6+\ldots +n^3\cdot 6\\ &=9+1^3\cdot 6+2^3\cdot 6+3^3\cdot 6+\ldots+n^3\cdot 6\\ & =9+\sum_{k=1}^n6k^3\\ &= \ldots\end{aligned}\]Can you take things from here? I hope this makes sense! :-)

Answer 2

@ChristopherToni .. yes i understood what you showed here, but the answer given is
(1/4) n (n+1)(n+4)(n+5) .. i am not getting this answer with the steps you have shown.

Answer 3

@ChristopherToni .. could you have a look at my previous comment :

yes i understood what you showed here, but the answer given is
(1/4) n (n+1)(n+4)(n+5) .. i am not getting this answer with the steps you have shown.