Value $\sum\limits_{r = 0}^{15} {\left( {{}^{15}{C_r}{}^{40}{C_{15}}{}^{20}{C_r} – {}^{35}{C_{15}}{}^{15}{C_r}{}^{25}{C_r}} \right)} $ is-

The coefficient of $x^{70}$ in $x^2(1+x)^{98}+x^3(1+x)^{97}+$ $x^4(1+x)^{96}+\ldots \ldots . .+x^{54}(1+x)^{46}$ is ${ }^{99} \mathrm{C}_p-{ }^{46} \mathrm{C}_{\mathrm{q}}$.

Then a possible value to $\mathrm{p}+\mathrm{q}$ is :

The sum, of the coefficients of the first $50$ terms in the binomial expansion of $(1-x)^{100}$, is equal to

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

If ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ………. + {C_n}{x^n}$, then $\frac{{{C_1}}}{{{C_0}}} + \frac{{2{C_2}}}{{{C_1}}} + \frac{{3{C_3}}}{{{C_2}}} + …. + \frac{{n{C_n}}}{{{C_{n – 1}}}} = $