If n>1, and (1+x)n=C0+C1x+C2x2+…+Cnxn then value of C0−2C1+3C2−4C3+…+(−1)n(n+1)Cn is

x(1+x)n=C0x+C1x2+C2x3+…+Cnxn+1 diff.w.r.t. x (1+x)n+nx(1+x)n-1=C0+C12x+C23x2+…+Cn(n+1)xnPutting x=-1 in the equation (1) of alternative solution we getC0−2C1+3C2−4C3+…+(−1)n(n+1)Cn=0+0=0

If n>1, and (1+x)n=C0+C1x+C2x2+…+Cnxn then value of C0−2C1+3C2−4C3+…+(−1)n(n+1)Cn is

x(1+x)n=C0x+C1x2+C2x3+…+Cnxn+1 diff.w.r.t. x (1+x)n+nx(1+x)n-1=C0+C12x+C23x2+…+Cn(n+1)xnPutting x=-1 in the equation (1) of alternative solution we getC0−2C1+3C2−4C3+…+(−1)n(n+1)Cn=0+0=0

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If $n > 1,$ and $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ then value of $C_{0} - 2 C_{1} + 3 C_{2} - 4 C_{3} + \dots + (- 1)^{n} (n + 1) C_{n}$ is

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Detailed Solution

$x (1 + x)^{n} = C_{0} x + C_{1} x^{2} + C_{2} x^{3} + \dots + C_{n} x^{n + 1} d i f f . w . r . t . x (1 + x)^{n} + n x {(1 + x)}^{n - 1} = C_{0} + C_{1} 2 x + C_{2} 3 x^{2} + \dots + C_{n} (n + 1) x^{n}$

Putting $x = - 1$ in the equation $(1)$ of alternative solution we get
$C_{0} - 2 C_{1} + 3 C_{2} - 4 C_{3} + \dots + (- 1)^{n} (n + 1) C_{n} = 0 + 0 = 0$