If Cr is coefficient of xr in expansion of 1-x n-n - N-n- 4 then -r-0n -1 r nCr 1-rln10 - 1-ln10n r is divisible remainder 0 by1-x n- n - N- n - 4 cr-xr - nr-0 -1 r n Cr 1-rln10 - 1-ln10n r-0

The correct options are A ^nC_2 B ^nC_4 C ^nC_1 D ^nC_3 S = ∑ _ r=0 ^ n ( 1)^ r ^ n C _ r #10; ( 1+rln10 ) #160; ( 1+nln10 ) ...

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Sum of Coefficients of All Terms

If Cr is co...

Question

If $C_{r}$ is coefficient of $x^{r}$ in expansion of ${(1 + x)}^{n}, n ϵ N, n \geq 4$ then $n \sum r = 0 {(- 1)}^{r} (^{n} C_{r}) \frac{(1 + r ln 10)}{{(1 + ln 10^{n})}^{r}}$ is divisible (remainder 0) by
(1+x) $^{n}$ , n $ϵ N$ , n $\geq 4$ $c_{r}, x^{r}$ $n \sum r = 0 {(- 1)}^{r} (^{n} C_{r}) \frac{(1 + r ln 10)}{{(1 + ln 10^{n})}^{r}} = 0$

^{n} C_{1}

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^{n} C_{3}

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^{n} C_{2}

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^{n} C_{4}

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Solution

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Similar questions

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

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

\sum_{r = 0}^{n - 1} {(\frac{{^{n} C}_{r}}{{^{n} C}_{r} + {^{n} C}_{r + 1}})}^{3} = \frac{4}{5}

n =

n \geq 1

C_{r} = (\begin{matrix} n r \end{matrix})

0 \leq r \leq n

{lim}_{x \to \infty} \sum_{r = 0}^{n} \frac{C_{r}}{(r + 3) n^{r}} = e - 2

\sum_{r = 0}^{n} \frac{C_{r}}{(r + 3) n^{r}} = \int_{0}^{1} {(1 + \frac{x}{n})}^{n} x^{2} d x

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\frac{P_{n + 1}}{P_{n}}

\frac{{(n + 1)}^{n}}{n!}

^{n + 1} C_{r + 1} = \frac{n + 1}{r + 1} .^{n} C_{r}

C_{r}

x^{r}

{(1 + x)}^{n}

n \sum r = 0 {(r + 1)}^{2} C_{r}

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