Binomial Theore...

If the coefficients of $$ (r+2)^{th} $$ and $$ (2r+1) ^{th}$$ terms $$ (r \neq 1) $$ are equal in the expansion of $$ (1+x)^{43} $$, then $$ r = $$

The middle term of $$ \left ( x - \dfrac{1}{x} \right )^{2n+1} $$ is

The ratio of the coefficient of $$ x^{10} $$ in $$ (1-x^2)^{10} $$ and  the term independent of $$x$$ in $$ \left ( x - \dfrac{2}{x} \right )^{10} $$ is

The $$3$$rd, $$4$$th and $$5$$th terms in the expansion of $$ (1+x)^n $$ are $$60, 160$$ and $$240$$ respectively, then $$ x = $$

Coefficient of $$ x^{16} $$ in $$ (1+x+x^2)(1-x)^{15} $$

The number of terms in the expansion of $$ (1+5\sqrt{2}x)^9 + (1-5\sqrt{2}x)^9 $$ is :

Coefficient of $$ x^5 $$ in $$ (1+x^2)^5(1+x)^4 $$ is

A. $$ ^{2n}C_n = C_0^2 + C_1^2 + C_2^2 + C_3^2 + \dots \dots + C_n^2 $$

B. $$ ^{2n}C_n = $$ term independent of $$x$$ in $$ (1+x)^n \left(1+\frac{1}{x} \right)^n $$

C. $$ ^{2n}C_n = \dfrac{1.3.5.7 \ldots \ldots (2n-1)}{n!} $$ then

The total number of terms in the expansion of $$ (x+a)^{100}+(x-a)^{100} $$ after simplification is

Coefficient of $$ x^{10} $$ in $$ (1+2x^4)(1-x)^8 $$ is