Question
If $\frac{1}{\sqrt{4 x+1}}\left\{\left(\frac{1+\sqrt{4 x+1}}{2}\right)^{n}-\left(\frac{1-\sqrt{4 x+1}}{2}\right)^{n}\right\}$$=a_{0}+a_{1} x+\ldots+a_{5} x^{5}$, then $n$ equals(A) 11(B) 9(C) 10(D) none of these
Step 1
Step 1: We are given the expression $\frac{1}{\sqrt{4 x+1}}\left\{\left(\frac{1+\sqrt{4 x+1}}{2}\right)^{n}-\left(\frac{1-\sqrt{4 x+1}}{2}\right)^{n}\right\}$ and we are asked to find the value of $n$ such that the expression can be written as a polynomial of Show more…
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$$ \begin{aligned} &\text { If }\\ &\begin{aligned} & \frac{1}{\sqrt{4 x+1}}\left[\left(\frac{1+\sqrt{4 x+1}}{2}\right)^{n}-\left(\frac{1-\sqrt{4 x+1}}{2}\right)^{n}\right] \\ =& a_{0}+a_{1} x+\cdots+a_{5} x^{5} \\ \text { then } n &-9 \mathrm{is} \end{aligned} \end{aligned} $$
$$ \begin{gathered} \frac{1}{\sqrt{2 x+1}}\left\{(1+\sqrt{2 x+1})^{n}-(1-\sqrt{2 x+1})^{n}\right\} \\ =a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{10} x^{10} \end{gathered} $$ then $n$ must be equal to (a) 20,21 (b) 21,22 (c) 22, 23 (d) 23, 24
If $n$ is a positive integer and $(1+x)^{n}=a_{0}+a_{1} x+a_{2} x^{2}+\ldots a_{n} x^{n}$, the value of $a_{0}-a_{2}+a_{4}-a_{6} \ldots . .-a_{102}$ (a) 0 (b) 1 (c) $-1$ (d) 2 $\sqrt{2}$
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