Sum to n terms of the series 11⋅2⋅3⋅4+12⋅3⋅4⋅5+13⋅4⋅5⋅6+⋯ is

The rth term of the series is given bytr=1r(r+1)(r+2)(r+3)Splitting tr into partial fraction, we gettr=16r−12(r+1)+12(r+2)−16(r+3)=161r−1r+1−131r+1−1r+2+161r+2−1r+3⇒∑r=1n tr=161−1n+1−1312−1n+2+1613−1n+3=n3+6n2+11n18(n+1)(n+2)(n+3)

Sum to n terms of the series 11⋅2⋅3⋅4+12⋅3⋅4⋅5+13⋅4⋅5⋅6+⋯ is

The rth term of the series is given bytr=1r(r+1)(r+2)(r+3)Splitting tr into partial fraction, we gettr=16r−12(r+1)+12(r+2)−16(r+3)=161r−1r+1−131r+1−1r+2+161r+2−1r+3⇒∑r=1n tr=161−1n+1−1312−1n+2+1613−1n+3=n3+6n2+11n18(n+1)(n+2)(n+3)

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Sum to n terms of the series $\frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \frac{1}{3 \cdot 4 \cdot 5 \cdot 6} + \dots$ is

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$\frac{n^{3} + 6 n^{2} + 11 n}{18 (n + 1) (n + 2) (n + 3)}$

$\frac{n^{3}}{3 (n + 1) (n + 2) (n + 3)}$

$\frac{15 n^{2} + 7 n}{4 n (n + 1) (n + 5)}$

$\frac{n^{2} + 6 n^{2} - 3 n}{6 (n + 2) (n + 3) (n + 4)}$

answer is D.

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Detailed Solution

The $rth$ term of the series is given by

$t_{r} = \frac{1}{r (r + 1) (r + 2) (r + 3)}$

Splitting $t_{r}$ into partial fraction, we get

$\begin{array}{l} t_{r} = \frac{1}{6 r} - \frac{1}{2 (r + 1)} + \frac{1}{2 (r + 2)} - \frac{1}{6 (r + 3)} \\ = \frac{1}{6} (\frac{1}{r} - \frac{1}{r + 1}) - \frac{1}{3} (\frac{1}{r + 1} - \frac{1}{r + 2}) + \frac{1}{6} (\frac{1}{r + 2} - \frac{1}{r + 3}) \\ \Rightarrow \sum_{r = 1}^{n} t_{r} = \frac{1}{6} (1 - \frac{1}{n + 1}) - \frac{1}{3} (\frac{1}{2} - \frac{1}{n + 2}) + \frac{1}{6} (\frac{1}{3} - \frac{1}{n + 3}) \\ = \frac{n^{3} + 6 n^{2} + 11 n}{18 (n + 1) (n + 2) (n + 3)} \end{array}$