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Sum of Coefficients of All Terms

If 1+x n= ∑...

Question

If ${(1 + x)}^{n} = n \sum r = 0 C_{r} x^{r}$ , then the value of the expression $a C_{0} + (a + b) C_{1} + (a + 2 b) C_{2} + \dots + (a + n b) C_{n}$ , equals

A

2 n a b (a + b) 2^{n}

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B

2 n a 2^{n + 1}

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C

(2 a + n b) 2^{n}

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D

(2 a + n b) 2^{n - 1}

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Solution

The correct option is D $(2 a + n b) 2^{n - 1}$
Simplifying we get
$a [^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + . . . .^{n} C_{n}] + b [^{n} C_{1} + 2^{n} C_{2} + 3^{n} C_{3} + . . . n^{n} C_{n}]$
$= a 2^{n} + b (\frac{d (1 + x)^{n}}{d x})_{x = 1}$
$= a 2^{n} + b (n (1 + x)^{n - 1})_{x = 1}$
$= a 2^{n} + b (n 2^{n - 1})$
$= (2 a + b n) 2^{n - 1}$

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Similar questions

a C_{0} + (a + b) C_{1} + (a + 2 b) C_{2} + . . . . + (a + n b) C_{n}

{(1 + x)}^{n} = n \sum r = 0 C_{r} x^{r}

, then the value of the expression

C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \frac{C_{3}}{4} + \dots + {(- 1)}^{n} \frac{C_{n}}{n + 1}

n \sum r = 0 C_{r} x^{r}

, then the value of

a C_{0} - (a + d) C_{1} + (a - 2 d) C_{2} - (a - 3 d) C_{3} + \dots + {(- 1)}^{n} (a - n d) C_{n}

{(1 + x)}^{n} = n \sum r = 0 C_{r} x^{r}

, then the value of

C_{0} - C_{2} + C_{4} - C_{6} + C_{8} - C_{10} + . . .

{(1 + x)}^{n} = n \sum r = 0 C_{r} x^{r}

the the value of

C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + . . . C_{n}^{2}

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