If 1-xn-C0-C1x-C2x2-Cnxn then show that n-xx-1x-2x-3-x-n-C0-x-C1-x-1-1nCn-x-n-

Let n! / x(x+1)(x+2)(x+3)...(x+n) = A_ 0 / x + A_ 1 / x+1 + A_ 2 / x+2 + A_ 3 / x+3 +...+ A_ n / x+n #160; ...(1)Therefore A_ 0 =lim _ x→ 0 x ...

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Byju's Answer

Standard XII

Mathematics

Expansions to Remove Indeterminate Form

If 1+xn=C0+...

Question

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}$ then show that $\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)} = \frac{C_{0}}{x} - \frac{C_{1}}{x + 1} - . . . + (- 1)^{n} \frac{C_{n}}{x + n}$ .

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Solution

Let $\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)} = \frac{A_{0}}{x} + \frac{A_{1}}{x + 1} + \frac{A_{2}}{x + 2} + \frac{A_{3}}{x + 3} + . . . + \frac{A_{n}}{x + n}$ ...(1)
Therefore
$A_{0} = lim x \to 0 x [\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)}] = lim x \to 0 [\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)}]$
$= \frac{n!}{1.2.3... n} = \frac{n!}{n!} = 1 =^{n} C_{0} = C_{0}$
$A_{1} = lim x \to - 1 (x + 1) [\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)}] = lim x \to - 1 [\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)}]$
$= \frac{n!}{(- 1) (1) (2) . . . (n - 1)} \times \frac{n}{n} = - \frac{n! n}{n!} = - n = -^{n} C_{1} = - C_{1}$
$A_{2} = lim x \to - 3 (x + 2) [\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)}] = lim x \to - 2 [\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)}]$
$= \frac{n!}{(- 2) (- 1) (1) . . . (n - 2)} = \frac{n (n - 1) (n - 2)!}{1 (n - 2)!} = \frac{n (n - 1)}{2} =^{n} C_{2} = C_{2}$
$A_{3} = lim x \to - 2 (x + 3) [\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)}] = lim x \to - 2 [\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)}]$
$= \frac{n!}{(- 3) (- 2) (- 1) (1) . . . (n - 3)} = \frac{n (n - 1) (n - 2) (n - 3)}{1.2.3 (n - 3)} = \frac{(n - 1) (n - 2)}{1.2.3} = -^{n} C_{3} = - C_{3}$
Similarly
$A_{n} = lim x \to n (x + n) [\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)}] = lim x \to - n [\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)}]$
$= \frac{n!}{(- n) (- n + 1) (- n + 2) (- n + 3) . . . (- 1)} = \frac{n!}{(- 1)^{n} [n (n - 1) (n - 2) (n - 3) . . .1]}$
$= \frac{(- 1)^{n} n!}{(- 1)^{2 n} n!} = (- 1)^{n}$
Substituting values of $A_{1}, A_{2}, A_{3}, . . ., A_{n}$ in (1) we get
$\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)} = \frac{C_{0}}{x} - \frac{C_{1}}{x + 1} + \frac{C_{2}}{x + 2} - \frac{C_{3}}{x + 3} + . . . + (- 1)^{n} C_{n} \frac{C_{n}}{x + n}$

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If (1+x)n=C0+C1x+C2x2+...+Cnxn then show that n!x(x+1)(x+2)(x+3)...(x+n)=C0x−C1x+1−...+(−1)nCnx+n.

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}$ then show that $\frac{n!}{x (x + 1) (x + 2) (x + 3) . . . (x + n)} = \frac{C_{0}}{x} - \frac{C_{1}}{x + 1} - . . . + (- 1)^{n} \frac{C_{n}}{x + n}$ .