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Find the sum of series 1.2.3 + 2.3.4 + 3.4.5 + ..... n terms

Option: 1

\frac{1}{4}[n(n+1)(n+2)(n-3)]


Option: 2

\frac{1}{4}[n(n+1)(n+2)(n+3)]


Option: 3

\frac{1}{3}[n(n+1)(n+2)(n+3)]


Option: 4

\frac{1}{3}[n(n-1)(n+2)(n-3)]


Answers (1)

best_answer

\\S_{n}=1 \cdot2 \cdot 3+2 \cdot3 \cdot 4+3 \cdot4 \cdot 5.... n\ terms\\ \\Here,\,T_{r}=r(r+1)(r+2)\\\text{Next factor is (r+3) and previous factor is (r-1)} \\(r+3)-(r-1)=4 \Rightarrow \frac{(r+3)-(r-1)}{4}=1 \\\text{Substitute this in }T_r\\T_{r}=\frac{1}{4}r(r+1)(r+2)[(r+3)-(r-1)]\\T_r=\frac{1}{4}[r(r+1)(r+2)(r+3)-(r-1)r(r+1)(r+2)]

So

\\T_1 = \frac{1}{4}[1.2.3.4 - 0.1.2.3] \\T_2 = \frac{1}{4}[2.3.4.5 - 1.2.3.4] \\T_3 = \frac{1}{4}[3.4.5.6 - 2.3.4.5] \\....\\....\\.... \\T_n = \frac{1}{4}[n(n+1)(n+2)(n+3) - (n-1)n(n+1)(n+2)]

Adding all these

\\S_n = \frac{1}{4}[n(n+1)(n+2)(n+3) - 0.1.2.3]\\ S_n = \frac{1}{4}n(n+1)(n+2)(n+3)

(For Vn Method students are advised NOT to remember the formula, but they should learn the method to find the sum)

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vinayak

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