JEE Main 2016 (22)

Algebra Level 4

$\dfrac{25!}{x(x+1)(x+2)\cdots(x+25)}=\sum_{i=0}^{25}\dfrac{A_{i}}{x+i}$

The equation above holds true for constants $A_1, A_2, \ldots , A_{25}$ . Find the sum of digits of the number $A_{24}$ .

Notation :
$!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

Try my set JEE Main 2016 .

7 None of these 13 11 6 9

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Mar 26, 2016

By analyzing the fractions made from $A = \frac{1}{x(x+1)(x+2)...(x+k)}$ we see that for $k = 2,$ $A = \frac{1}{1 * 2 * x} + \frac{1}{-1 * 1 * (x+1)} + \frac{1}{-2 * -1 * (x+2)}$

Continuing for larger values of k, using the cover-up method for partial fractions, we can develop the following summation to generalize for any integer $k > 0,$

$A = \frac{1}{x(x+1)(x+2)...(x+k)} = \sum_{i=0}^k \frac{(-1)^i}{(x+i)(i!)(k-i)!}$

All we need is the $25^{th}$ fraction corresponding to $i = 24$ . $\frac{25!*(-1)^{24}}{(x+24)(24!)(25-24)!}$ which reduces to $\boxed{\frac{25}{x+24}}$

JEE Main 2016 (22)

Try my set JEE Main 2016 .

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