${1.2}^{2} + {2.3}^{2} + {3.4}^{2} + . . . .$ upto n terms, is equal to -

\frac{1}{12} n (n + 1) (n + 2) (n + 3)

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\frac{1}{12} n (n + 1) (n + 2) (n + 5)

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\frac{1}{12} n (n + 1) (n + 2) (3 n + 5)

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None of these

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Solution

The correct option is A $\frac{1}{12} n (n + 1) (n + 2) (3 n + 5)$
Here $T_{n} = n (n + 1)^{2}$

$∴ S_{n} = \sum T_{n} = \sum n^{3} + 2 \sum n^{2} + \sum n$

$= \frac{n^{2} (n + 1)^{2}}{4} + 2. \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}$

$= \frac{1}{12} n (n + 1) (n + 2) (3 n + 5)$

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Similar questions

n^{th}

n

2, 12, 36, 80, 150, 252, . . . . .

T_{n}

S_{n}

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

= (- 1)^{n + 1} \frac{(2 n)}{2 (n!)^{2}},

n \sum r = 1 t_{r} = \frac{1}{12} n (n + 1) (n + 2)

n \sum r = 1 \frac{1}{t_{r}}

n \sum r = 1 t_{r} = \frac{1}{12} n (n + 1) (n + 2)

n \sum r = 1 \frac{1}{t_{r}}

n

\frac{C_{0}}{1 \cdot 2} + \frac{C_{1}}{2 \cdot 3} + \frac{C_{2}}{3 \cdot 4} + \dots

C_{r} =^{n} C_{r}

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