If $3x = 1 + \frac{5}{8} + \frac{5 \times 9}{8 \times 16} + \frac{5 \times 9 \times 13}{8 \times 16 \times 24} + \dots$,then $x^4 + 4x^3 + 6x^2 + 4x = $

  • A
    $0$
  • B
    $1$
  • C
    $4$
  • D
    $8$

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Similar Questions

When $|x|>3$,the coefficient of $\frac{1}{x^n}$ in the expansion of $x^{3/2}(3+x)^{1/2}$ is

$1+\frac{2}{4}+\frac{2 \cdot 5}{4 \cdot 8}+\frac{2 \cdot 5 \cdot 8}{4 \cdot 8 \cdot 12}+\frac{2 \cdot 5 \cdot 8 \cdot 11}{4 \cdot 8 \cdot 12 \cdot 16}+\ldots \ldots$ is equal to :

$\frac{1}{(2 + x)^4} = $

The sum of the infinite series $1+\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots$ is equal to

If $x = \frac{3}{4 \cdot 8} + \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} + \ldots$,then $2x^2 + 5x =$

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