The coefficient of xn in the expansion of 1-3x-6x2-10x3-1rr-1r-2xr-2- -n is

The correct option is A 3n!/n! 2n! Given, (1 3x+6x^2+...∞)^ n =[(1+x)^ 3]^ n =(1+x)^3n Hence, #160;Coefficient of x^n will be #16 ...

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Logarithmic Function

The coefficie...

Question

The coefficient of $x^{n}$ in the expansion of $(1 - 3 x + 6 x^{2} - 10 x^{3} + . . . + (- 1)^{r} \frac{(r + 1) (r + 2) x^{r}}{2} + . . . \infty)^{- n}$ is

\frac{3 n!}{(n!)^{2}}

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\frac{3 n!}{n! 2 n!}

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\frac{3 n!}{(2 n!)^{2}}

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None of these

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x^{r} [0 \leq r \leq (n - 1)]

{(x + 3)}^{n - 1} + {(x + 3)}^{n - 2} (x + 2) + {(x + 3)}^{n - 3} {(x + 2)}^{2} + . . . . . + {(x + 2)}^{n - 1}

The (n+1)th term from the end in the expansion of $(2 x - \frac{1}{x})^{3 n}$ is:

x^{r} (0 \leq r \leq (n - 1))

(x + 3)^{n - 1} + (x + 3)^{n - 2} (x + 2) + (x + 3)^{n - 3} (x + 2)^{2} + \dots + (x + 2)^{n - 1}

x^{n - 1}

(x + 3)^{n} + (x + 3)^{n - 1} (x + 2) + (x + 3)^{n - 2} (x + 2)^{2} + . . . . . . . + (x + 2)^{n}

x

{(x - \frac{1}{x^{2}})}^{3 n}

n

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