Find the sum to n terms of the series 1-1-2-1-2-3-1-2-3-4- -A- nn-1-2B- nn-1n-2-6C- nn-12 n-1-6D- nn-1n-2-12

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Divisibility Rule for Composite Numbers

Find the sum ...

Question

Find the sum to n terms of the series 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ...............................................:

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n

(1 \times 2 \times 3) + (2 \times 3 \times 4) + (3 \times 4 \times 5) + \dots

1 \times n + 2 (n - 1) + 3 (n - 2) + \dots \dots + (n - 1) \times 2 + n \times 1

n

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

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

n

1^{n} - n 2^{n} + \frac{n (n - 1)}{1 \cdot 2} 3^{n} - \dots = {(- 1)}^{n} n!

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

n + 1

If 1+ $\frac{1 + 2}{2} + \frac{1 + 2 + 3}{3} + . . . .$ to n terms is S. Then, S is equal to

\frac{n (1 + x)}{1 + n x} - \frac{n (n - 1)}{⌊ 2} \cdot \frac{1 + 2 x}{(1 + n x)^{2}} + \frac{n (n + 1) (n - 2)}{⌊ 3} \cdot \frac{1 + 3 x}{(1 + n x)^{3}} - . . . .

n

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