If ${\rm{\alpha }} = \frac{5}{{2!3}} + \frac{{5.7}}{{3!{3^2}}} + \frac{{5.7.9}}{{4!{3^2}}} + . \ldots \ldots ..,\;$ if then ${{\rm{\alpha }}^2} + 4{\rm{\alpha }}$ equal to

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Medium

1 of 1

Analyzing +4 | -1

\mathrm{\alpha } = \frac{5}{{2!3}} + \frac{{5.7}}{{3!{3^2}}} + \frac{{5.7.9}}{{4!{3^2}}} + . \ldots \ldots ..,\;

if then

{\mathrm{\alpha }^2} + 4\mathrm{\alpha }

equal to

View answer and explanation

Correct option: B

Option B (23) is correct because, after comparing the given series with the binomial expansion and solving for α, we find that α^2 + 4α equals 23.

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