Let $f(x)=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 x}$ and $g(x)=b_0+b_1 x+b_2 x^2 +\ldots+b_{n-1} x^{n-1} +x^n+x^{n+1}+\ldots+x^{2 n}$ If $f(x)=g(x+1)$, then $a_n$ in terms of $n$ is equal to

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Medium

1 of 1

Analyzing +4 | -1

Let

f(x)=a_0+a_1 x+a_2 x^2+\ldots+a_{2 n} x^{2 x}

and

g(x)=b_0+b_1 x+b_2 x^2 +\ldots+b_{n-1} x^{n-1} +x^n+x^{n+1}+\ldots+x^{2 n}

f(x)=g(x+1)

, then

a_n

in terms of

n

is equal to

View answer and explanation

Correct option: A

{ }^{2 n+1} C_{n+1}

Option B (

{ }^{2 n+1} C_{n+1}

) is correct because it represents the coefficient of

x^n

after transforming the given polynomial expression using binomial theorem.

Practice All Chapter MCQs →

\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n