# How to solve find the sum to n terms of the series 1*2*3 + 2*3*4

Welcome to my article How to solve find the sum to n terms of the series 1*2*3+2*3*4. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve find the sum to n terms of the series 1*2*3+2*3*4, read and understand it carefully till the end.

Let us know how to solve this question How to solve find the sum to n terms of the series 1*2*3+2*3*4.

First write the question on the page of the notebook.

## How to solve find the sum to n terms of the series 1*2*3+2*3*4

First write this question as follows and then solve,

\displaystyle \left( {1\times 2\times 3} \right)+\left( {2\times 3\times 4} \right)+\left( {3\times 4\times 5} \right)+……..n

\displaystyle \because Tn=n\left[ {\left( {n+1} \right)\left( {n+2} \right)} \right]

\displaystyle Tn=n\left[ {{{n}^{2}}+3n+2} \right]

\displaystyle Tn={{n}^{3}}+3{{n}^{2}}+2n

\displaystyle \sum{{Tn}}=\sum{{({{n}^{3}}}}+3{{n}^{2}}+2n)

\displaystyle \sum{{Tn}}=\sum{{{{n}^{3}}}}+3\sum{{{{n}^{2}}}}+2\sum{n})

\displaystyle \sum{{Tn}}={{\left[ {\frac{{n(n+1)}}{2}} \right]}^{2}}+3\left[ {\frac{{n(n+1)(2n+1)}}{6}} \right]+2\left[ {\frac{{n(n+1)}}{2}} \right]

\displaystyle \sum{{Tn}}=\frac{{{{n}^{2}}{{{(n+1)}}^{2}}}}{4}+\frac{3}{6}\left[ {({{n}^{2}}+n)(2n+1)} \right]+\frac{2}{2}\left[ {n(n+1)} \right]

\displaystyle \sum{{Tn}}=\frac{{{{n}^{2}}({{n}^{2}}+2n+1)}}{4}+\frac{1}{2}(2{{n}^{3}}+{{n}^{2}}+2{{n}^{3}}+n)+({{n}^{2}}+n)

\displaystyle \sum{{Tn}}=\frac{{({{n}^{4}}+2{{n}^{3}}+{{n}^{2}})}}{4}+\frac{{(2{{n}^{3}}+{{n}^{2}}+2{{n}^{3}}+n)}}{2}+\frac{{({{n}^{2}}+n)}}{1}

\displaystyle \sum{{Tn}}=\frac{{({{n}^{4}}+2{{n}^{3}}+{{n}^{2}})}}{4}+\frac{{2\times (2{{n}^{3}}+{{n}^{2}}+2{{n}^{3}}+n)}}{{2\times 2}}+\frac{{4\times ({{n}^{2}}+n)}}{{1\times 4}}

\displaystyle \sum{{Tn}}=\frac{{({{n}^{4}}+2{{n}^{3}}+{{n}^{2}})}}{4}+\frac{{(4{{n}^{3}}+2{{n}^{2}}+4{{n}^{3}}+2n)}}{4}+\frac{{(4{{n}^{2}}+4n)}}{4}

\displaystyle \sum{{Tn}}=\frac{{({{n}^{4}}+2{{n}^{3}}+{{n}^{2}})+(4{{n}^{3}}+2{{n}^{2}}+4{{n}^{3}}+2n)+(4{{n}^{2}}+4n)}}{4}

\displaystyle \sum{{Tn}}=\frac{{{{n}^{4}}+2{{n}^{3}}+{{n}^{2}}+8{{n}^{3}}+2{{n}^{2}}+2n+4{{n}^{2}}+4n}}{4}

\displaystyle \sum{{Tn}}=\frac{{{{n}^{4}}+10{{n}^{3}}+7{{n}^{2}}+6n}}{4}

\displaystyle \sum{{Tn}}=\frac{n}{4}\left[ {{{n}^{3}}+10{{n}^{2}}+7n+6} \right]

so that