If $(x + 1) (x + 2) . . . (x + n) = A_{0} + A_{1} x + . . . . A_{n} x^{n}$ then, $A_{1} + 2 A_{2} . . . n A_{n}$ is equal to:

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

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

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

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None of the above.

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Solution

The correct option is D $(n + 1)! \times (\frac{1}{2} + \frac{1}{3} . . . . . \frac{1}{n + 1})$
Consider the summation $A_{0} + A_{1} x + A_{2} x^{2} + . . . + A_{n} x^{n 0}$
Now,differentiating the summation with respect to $x$ and substituting $x = 1$ gives,
$A_{0} + A_{1} + {2 A}_{2} + . . . + n A_{n}$ ,
$∴$ To get the value of the required summation we need to differentiatie the $L . H . S$ with respect to $x$ and we need to substitute $x = 1$ ,
Let $y = (x + 1) (x + 2) (x + 3) . . . (x + n)$ ,
applying $l o g a r i t h m$ and differentiating with respect to $x$
$\frac{d y}{d x} = ((x + 2) (x + 3) . . . (x + n)) + ((x + 1) (x + 3) . . . (x + n)) + . . . + ((x + 1) (x + 2) . . . (x + n - 1))$
Substituting $x = 1$ gives,
$(n + 1)! \times (\frac{1}{2} + \frac{1}{3} + . . . \frac{1}{n})$

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