JEE MATHEMATICS MCQ: BINOMIAL THEOREM

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Application +4 | -1

If the polynomial

f(x)=1-x+x^2-x^3+\cdots+x^{19}+x^{20}

is expressed as

g(y)=a_0+a_1 y+a_2 y^2+\cdots+a_{20} y^{20}

, where

y=x-4

, then the value of

a_0+a_1+a_2+\cdots+a_{20}

View answer and explanation

Correct option: D

\frac{1+5^{21}}{6}

Option D (

\frac{{1+5^{21}}}{{6}}

) is correct. The value of the polynomial is found by substituting y=1 into the expression g(y) after converting f(x) to a geometric series.

More Questions from 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