Question

The sum of first 10 terms of the series $1\cdot 2\cdot 3+2\cdot 3\cdot 4+3\cdot 4\cdot 5+.........$ is

• A. 6006
• B. 2970
• C. 3990
• D. 4290

Answer 1

The correct option is D 4290

$\mathrm{The}\mathrm{nth}\mathrm{term}\mathrm{of}\mathrm{the}\mathrm{series}\mathrm{is}\mathrm{given}\mathrm{by}{T}_n=n\left(n+1\right)\left(n+2\right)\mathrm{If}{T}_n\mathrm{is}\mathrm{multiplication}\mathrm{of}n\mathrm{consecutive}\mathrm{intergers},\mathrm{then}\mathrm{we}\mathrm{take}{V}_n\mathrm{as}\mathrm{multiplication}\mathrm{of}n+1\mathrm{consecutive}\mathrm{integers}.\mathrm{Let}{V}_n=n\left(n+1\right)\left(n+2\right)\left(n+3\right)V_n-V_{n-1}=\left(n\left(n+1\right)\left(n+2\right)\left(n+3\right)\right]-\left((n-1)\left(n-1+1\right)\left(n-1+2\right)\left(n-1+3\right)\right]\Rightarrow V_n-V_{n-1}=n\left(n+1\right)\left(n+2\right)\left(\left(n+3\right)-(n-1)\right]\Rightarrow V_n-V_{n-1}=4n\left(n+1\right)\left(n+2\right)\Rightarrow \frac{V_n-V_{n-1}}{4}=n\left(n+1\right)\left(n+2\right)\Rightarrow \frac{V_n-V_{n-1}}{4}=T_n\mathrm{Sum}\mathrm{of}n\mathrm{tems}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{sequence}\mathrm{will}\mathrm{be}{S}_n={\sum }_{k=1}^n{T}_n\Rightarrow S_n={\sum }_{k=1}^n\frac{\left(}{V_n-V_{n-1}}\Rightarrow S_n=\frac{1}{4}\left(V_1-V_0+V_2-V_1+V_3-V_2+.....+V_n-V_{n-1}\right)\Rightarrow S_n=\frac{1}{4}\left(V_n-V_0\right)\Rightarrow S_n=\frac{1}{4}{V}_n(\mathrm{As}{V}_0=0)\Rightarrow S_n=\frac{1}{4}n\left(n+1\right)\left(n+2\right)\left(n+3\right)\mathrm{For}n=10S_{10}=\frac{1}{4}10\left(10+1\right)\left(10+2\right)\left(10+3\right)\Rightarrow S_{10}=4290$