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Differentiation to Solve Modified Sum of Binomial Coefficients

If 1+x n=∑r...

Question

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

A

n 2^{n}

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B

(n + 1) 2^{n + 1}

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C

2^{n + 1}

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D

n 2^{n - 1}

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Solution

The correct option is D $n 2^{n - 1}$
Simplifying, we get
$n^{n} C_{0} + (n - 1)^{n} C_{1} + (n - 2)^{n} C_{2} + (n - 3)^{n} C_{3} + (n - 4)^{n} C_{4} . . . (n - n)^{n} C_{n}$
$= n (^{n} C_{0} +^{n} C_{1} +^{n} C_{2} + . . .^{n} C_{n}) - (^{n} C_{1} + 2^{n} C_{2} + 3^{n} C_{3} + . . . n^{n} C_{n})$
$= [n (1 + x)^{n} + \frac{d (1 + x)^{n}}{d x}] |_{x = 1}$
$= n 2^{n} + n 2^{n - 1}$
$= 2^{n - 1} (n) [2 - 1]$
$= n 2^{n - 1}$

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Similar questions

C_{0}, C_{1}, C_{2}, . . . . . . . . . . . C_{n}

are the Binomial coefficients in the expansion

(1 + x)^{n} .

‘n’ being even, then

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . . . . . (C_{0} + C_{1} + C_{2} + . . . . . + C_{n - 1})

C_{0}, C_{1}, C_{2} . . . . . . . . . . ., C_{n}

are the binomial coefficient in the expansion of

(1 + x)^{n}, n

being even, then

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . . . + (C_{0} + C_{1} + C_{2} + . . . . . . . . + C_{n - 1})

C_{0}, C_{1}, C_{2}, . . . . . . . . . . C_{n}

are the binomial coefficients in the expansion of

(1 + x)^{n}

. n being even, then

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . . . . + (C_{0} + C_{1} + C_{2} + . . . . . . . . . + C_{n - 1})

{(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . . . . + C_{n} x^{R}

C_{0} + (C_{0} + C_{1}) + (C_{0} + C_{1} + C_{2}) + . . . . . + (C_{0} + C_{1} + C_{2} + . . . . . + C_{n - 1}

c_{0}, c_{1}, c_{2}, . . . . c_{n}

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(1 + x)^{n}

c_{0} + \frac{c_{1}}{2} + \frac{c_{2}}{3} + . . . . . + \frac{c_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}

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