$C_{0} + 2 C_{1} + 3. C_{2} + . . . + (n + 1) C_{n} = 2^{n - 1} (n + 2) .$

True

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False

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Solution

The correct option is A True
${(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}$
MUltiplying both sides by $X,$
$x {(1 + x)}^{n} = C_{0} x + C_{1} x^{2} + C_{2} x^{3} + . . . + C_{n} x^{n + 1}$
Differentiating both sides
${(1 + x)}^{n} + x n {(1 + x)}^{n - 1} = C_{0} + 2. C_{1} x + 3. C_{2} x^{2} + . . . + (n + 1) C_{n} x^{n}$
putting $x = 1,$ we get
$C_{0} + 2. C_{1} + 3. C_{2} + . . . + (n + 1) C_{n} = 2^{n} + n {.2}^{n - 1}$
$C_{0} + 2. C_{1} + 3. C_{2} + . . . + (n + 1) C_{n} = 2^{n - 1} (n + 2)$

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