If 2x+3(x+1)4=Ax+1+B(x+1)2+C(x+1)3+D(x+1)4, then A+B+C+D=

Let x+1=a⇒x=a−1∴2x+3(x+1)4=2(a−1)+3a4=2a+1a4=2aa4+1a4=2a3+1a4∴A+B+C+D=0+0+2+1=3

If 2x+3(x+1)4=Ax+1+B(x+1)2+C(x+1)3+D(x+1)4, then A+B+C+D=

Let x+1=a⇒x=a−1∴2x+3(x+1)4=2(a−1)+3a4=2a+1a4=2aa4+1a4=2a3+1a4∴A+B+C+D=0+0+2+1=3

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If $\frac{2 x + 3}{{(x + 1)}^{4}} = \frac{A}{x + 1} + \frac{B}{{(x + 1)}^{2}} + \frac{C}{{(x + 1)}^{3}} + \frac{D}{{(x + 1)}^{4}}$ , then $A + B + C + D =$

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answer is D.

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Let $x + 1 = a \Rightarrow x = a - 1$

$∴ \frac{2 x + 3}{{(x + 1)}^{4}} = \frac{2 (a - 1) + 3}{a^{4}} = \frac{2 a + 1}{a^{4}} = \frac{2 a}{a^{4}} + \frac{1}{a^{4}} = \frac{2}{a^{3}} + \frac{1}{a^{4}}$

$∴ A + B + C + D = 0 + 0 + 2 + 1 = 3$

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If 2x+3(x+1)4=Ax+1+B(x+1)2+C(x+1)3+D(x+1)4, then A+B+C+D=

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If $\frac{2 x + 3}{{(x + 1)}^{4}} = \frac{A}{x + 1} + \frac{B}{{(x + 1)}^{2}} + \frac{C}{{(x + 1)}^{3}} + \frac{D}{{(x + 1)}^{4}}$ , then $A + B + C + D =$