If e^ x +e^ 5 x e^ 3 x =a_ 0 +a_ 1 x+a_ 2 x^ 2 +a_ 3 x^ 3 + then the value of 2 a_ 1 +2^ 3 a_ 3 +2^ 5 a_ 5 + is

Correct answer: D. 0 Let ^ x + ^ 5 x ^ 3 x =a_ 0 +a_ 1 x+a_ 2 ^ 2 +a_ 3 ^ 3 + . aligned &= ^ x ^ 3 x + ^ 5 x ^ 3 x =a_ 0 +a_ 1 x+a_ 2 ^ 2 + . =& ^ -2 x + ^ 2 x =a_ 0 +a_ 1 x+a_ 2 ^ 2 +a_ 3 ^ 3 + . aligned By using ^ x =1+x+ ^ 2 2 ! + ^ 3 3 ! + _ -- -- and ^ -x =1-x+ ^ 2 2 ! - ^ 3 3 ! +--- ^ -2 x + ^ 2 x =2 [1+ (2 x)^ 2 21 + (2 x)^ 4 41 + .] array l =a_ 0 +a_ 1 x+a_ 2 x^ 2 +a_ 3 a^ 3 + . =a_ 1 =a_ 3 =a_ 5 = =0 array Hence, 2 a_ 1 +2^ 3 a_ 3 +2^ 5 a_ 5 + .=0

BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Mathematics
Chapter: Binomial Theorem

BITSAT 2015 · Mathematics · Binomial Theorem

If e^ x +e^ 5 x e^ 3 x =a_ 0 +a_ 1 x+a_ 2 x^ 2 +a_ 3 x^ 3 + then...

\frac{e^{x}+e^{5 x}}{e^{3 x}}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\ldots

then the value of

2 a_{1}+2^{3} a_{3}+2^{5} a_{5}+\ldots

Options

A $e^{2}+e^{-2}$
B $e^{4}-e^{-4}$
C $e^{t}+e^{-1}$
D0

Correct answer

D. 0

Step-by-step solution

Let

\frac{\mathrme^{\mathrm{x}}+\mathrme^{5 \mathrm{x}}}{\mathrme^{3 \mathrm{x}}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrmx^{2}+\mathrm{a}_{3} \mathrma^{3}+\ldots.

\begin{aligned} &=\frac{\mathrme^{\mathrm{x}}}{\mathrme^{3 \mathrm{x}}}+\frac{\mathrme^{5 \mathrm{x}}}{\mathrme^{3 \mathrm{x}}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrmx^{2}+\ldots. \\=& \mathrme^{-2 \mathrm{x}}+\mathrme^{2 \mathrm{x}}=\mathrm{a}_{0}+\mathrm{a}_{1} \mathrm{x}+\mathrm{a}_{2} \mathrmx^{2}+\mathrm{a}_{3} \mathrmx^{3}+\ldots. \end{aligned}

By using $\mathrme^{\mathrm{x}}=1+\mathrm{x}+\frac{\mathrmx^{2}}{2 !}+\frac{\mathrmx^{3}}{3 !}+{ }_{--}--$ and $\mathrme^{-\mathrm{x}}=1-\mathrm{x}+\frac{\mathrmx^{2}}{2 !}-\frac{\mathrmx^{3}}{3 !}+---\mathrme^{-2 \mathrm{x}}+\mathrme^{2 \mathrm{x}}=2 \left[1+\frac{(2 \mathrm{x})^{2}}{21}+\frac{(2 \mathrm{x})^{4}}{41}+\ldots.\right]

\begin{array}{l} =a_{0}+a_{1} x+a_{2} x^{2}+a_{3} a^{3}+\ldots. \\ =a_{1}=a_{3}=a_{5}=\ldots=0 \end{array}$

Hence, 2 \mathrm{a}_{1}+2^{3} \mathrm{a}_{3}+2^{5} \mathrm{a}_{5}+\ldots.=0

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