Question

# Find the $$nth$$ term and the sum of $$n$$ terms of the series $$1.2.4+2.3.5+3.4.6+....$$

Solution

## $$1.2.4+2.3.5+3.4.6+….$$$$\therefore n^{th} term= n\left(n+1\right)\left(n+3\right)$$$$\therefore$$ sum $$= \displaystyle \sum_{k=1}^{n}{k\left(k+1\right) \left(k+3\right)}$$$$= \displaystyle \sum_{k=1}^{n}\left({k}^{3}+4{k}^{2}+3k\right)$$$$= \displaystyle \sum_{k=1}^{n}{{k}^{3}}+4\sum_{k=1}^{n}{{k}^{2}}+3\sum_{k=1}^{n}{k}$$$${\left[\dfrac{n\left(n+1\right)}{2}\right]}^{2}+\dfrac{4n\left(n+1\right) \left(2n+1\right)}{6}+3\dfrac{n\left(n+1\right)}{2}$$$$=\dfrac{{n}^{2}{\left(n+1\right)}^{2}}{4}+\dfrac{2n\left(n+1\right)\left(2n+1\right)}{3}+\dfrac{3}{2}n\left(n+1\right)$$$$=n\left(n+1\right) \left[\dfrac{n\left(n+1\right)}{4}+\dfrac{n\left(2n+1\right)}{3}\dfrac{3}{2}\right]$$$$=n\left(n+1\right)\left[\dfrac{3{n}^{2}+3n+16n+8+18}{12}\right]$$$$=\dfrac{n\left(n+1\right)+ \left(3{n}^{2}+19n+26\right)}{12}$$ AnsMathematics

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