Binomial Theorem — Page 16 | JEE Mathematics

JEE Mathematics · 151 questions · Page 16 of 16 · Click an option or "Show Solution" to reveal answer

Q151

Let Sn = 1 + q + q2 + . . . . . + qn and Tn = 1 +

\left( {{{q + 1} \over 2}} \right) + {\left( {{{q + 1} \over 2}} \right)^2}

+ . . . . . .+

{\left( {{{q + 1} \over 2}} \right)^n}

where q is a real number and q

\ne

1. If 101C1 + 101C2 . S1 + .... + 101C101 . S100 =

\alpha

T100 then

\alpha

is equal to

A 202

B 200

C 2100

D 299

Correct Answer

Option C

Solution

101C1 + 101C2S1 + . . . . . . .

+ 101C101S100 $=$ $\alpha$ T100 101C1 + 101C2(1 + q) + 101C3(1 + q + q2) + . . . . . .

+101C101(1 + q + . . . . . + q100)

= 2\alpha {{\left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)} \over {\left( {1 - q} \right)}}

$\Rightarrow$ 101C1(1 $-$ q) + 101C2(1 $-$ q2) + . . . . . . + 101C101(1 $-$ q101)

= 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)

$\Rightarrow$ (2101 $-$ 1) $-$ ((1 + q)101 $-$ 1)

= 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)

$\Rightarrow$

{2^{101}}\left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right) = 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)

$\Rightarrow$

\alpha = {2^{100}}