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Formation of a Differential Equation from a General Solution

If 1+xn=C0+...

Question

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . + C_{n} x^{n}$ , then the value of $C_{0} + 2 C_{1} + 3 C_{2} + . . . . + (n + 1) C_{n}$ will be

A

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

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B

(n + 1) 2^{n}

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C

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

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D

(n + 2) 2^{n}

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Solution

The correct option is A $(n + 2) 2^{n - 1}$
Since we know that $2^{n} = C_{0}^{n} + C_{1}^{n} + C_{2}^{n} {+ C_{3}^{n} + C}_{4}^{n} + C_{5}^{n} + . . . . = \sum_{r = o}^{r = n} C_{r}^{n} C_{0}^{n} + {2 C}_{1}^{n} + {3 C}_{2}^{n} {+ {4 C}_{3}^{n} + 5 C}_{4}^{n} + {6 C}_{5}^{n} + . . . . = \sum_{r = 0}^{n} {(r + 1) C}_{r}^{n} = \sum_{r = 0}^{n} {(r) C}_{r}^{n} + \sum_{r = o}^{r = n} C_{r}^{n} ⟹ {r C}_{r}^{n} = {n C}_{r - 1}^{n - 1} ∴ \sum_{r = 0}^{n} {(r + 1) C}_{r}^{n} = n 2^{n - 1} + 2^{n} = (n + 2) 2^{n - 1}$

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Similar questions

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},

then
the value of

\sum \sum 0 \leq r < s \leq n (r + s) C_{r} C_{s}

C_{0} + 2 C_{1} + 3 C_{2} + 4 C_{3} + . . . + (n + 1) C_{n} = (n + 2) 2^{n - 1}

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

C_{0} + 2 C_{1} + 3 C_{2} + . . . . . . . . . + (n + 1) C_{n}

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

3 C_{0} + 3^{2} \frac{C_{1}}{2}

\frac{3^{3} C_{2}}{3} + \frac{3^{4} C_{3}}{4}

\frac{3^{n + 1} C_{n}}{n + 1}

\frac{4^{n + 1} - 1}{n + 1}

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n},

then the value of

\sum \sum 0 \leq r < s \leq n {(C_{r} + C_{s})}_{r}^{2}

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