In the expansion of $(1 + x)^5$,the sum of the coefficients of the terms is

  • A
    $80$
  • B
    $16$
  • C
    $32$
  • D
    $64$

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If $b$ is very small as compared to the value of $a$,so that the cube and other higher powers of $\frac{b}{a}$ can be neglected in the identity $\frac{1}{a-b}+\frac{1}{a-2b}+\frac{1}{a-3b}+\ldots+\frac{1}{a-nb}=\alpha n+\beta n^2+\gamma n^3$,then the value of $\gamma$ is:

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