The Value of the sum 1.2.3+2.3.4+3.4.5+....upto n terms =

\frac{1}{6} n^{2} (2 n^{2} + 1)

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\frac{1}{6} (n^{2} - 1) (2 n - 1) (2 n + 3)

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\frac{1}{8} (n^{2} + 1) (n^{2} + 5)

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

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Solution

The correct option is D $\frac{1}{4} n (n + 1) (n + 2) (n + 3)$
1.2.3+2.3.4+.......+ upto n terms
$t_{n} = n (n + 1) (n + 2)$
$S_{n} = \sum t_{n} = \frac{1}{4} n (n + 1) (n + 2) (n + 3)$

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Similar questions

n

\frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} + . . . . .

n

\frac{1}{1.2.3} + \frac{3}{2.3.4} + \frac{5}{3.4.5} + \frac{7}{4.5.6} + . . . .

\frac{4}{1.2.3} \cdot \frac{1}{3} + \frac{5}{2.3.4} \cdot \frac{1}{3^{2}} + \frac{6}{3.4.5} \cdot \frac{1}{3^{3}} + . . . .

$\frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} + \dots + \frac{1}{n (n + 1) (n + 2)} = \frac{n (n + 3)}{4 (n + 1) (n + 2)} .$

\frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} + . . . . . + \frac{1}{n (n + 1) (n + 2)} = \frac{n (n + 3)}{4 (n + 1) (n + 2)}

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