Find an approximation of $(0.99)^{5}$ using the first three terms of its expansion.
$0.99=1-0.01$
$\therefore(0.99)^{5}=(1-0.01)^{5}$
$ = {\,^5}{C_0}{(1)^5} - {\,^5}{C_1}{(1)^4}(0.01) + {\,^5}{C_2}{(1)^3}{(0.01)^2}$ [ Approximately ]
$=1-5(0.01)+10(0.01)^{2}$
$=1-0.05+0.001$
$=1.001-0.05$
$=0.951$
Thus, the value of $(0.99)^{5}$ is approximately $0.951$
The middle term in the expansion of ${\left( {x + \frac{1}{x}} \right)^{10}}$ is
In the binomial expansion of ${\left( {a - b} \right)^n},n \ge 5,\;$ the sum of $5^{th}$ and $6^{th}$ terms is zero , then $a/b$ equals.
Find the coefficient of $x^{5}$ in the product $(1+2 x)^{6}(1-x)^{7}$ using binomial theorem.
The term independent of $x$ in the expansion of $\left[\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right]^{10}, x \neq 1,$ is equal to ....... .
The number of terms in the expansion of ${\left( {\sqrt[4]{9} + \sqrt[6]{8}} \right)^{500}}$, which are integers is