WBJEE · Maths · Sequences and Series
The coefficient of \(x^n\) in the expansion of \(\frac{e^{7 x}+e^x}{e^{3 x}}\) is
- A \(\frac{4^{\mathrm{n}-1}-(-2)^{\mathrm{n}-1}}{\lfloor\mathrm{n}}\)
- B \(\frac{4^{n-1}-2^{n-1}}{\lfloor\text { n }}\)
- C \(\frac{4^{\mathrm{n}}-2^{\mathrm{n}}}{\lfloor\mathrm{n}}\)
- D \(\frac{4^{\mathrm{n}}+(-2)^{\mathrm{n}}}{\underline{n}}\)
Answer & Solution
Correct Answer
(D) \(\frac{4^{\mathrm{n}}+(-2)^{\mathrm{n}}}{\underline{n}}\)
Step-by-step Solution
Detailed explanation
Hints: \(\frac{e^{7 x}+e^x}{e^{3 x}}=e^{4 x}+e^{-2 x}\) Co-efficient of \(x^n\) \(\frac{(4)^{\mathrm{n}}}{\mathrm{n} !}+\frac{(2)^{\mathrm{n}}}{\mathrm{n} !}(-1)^{\mathrm{n}}=\frac{4^{\mathrm{n}}+(-2)^{\mathrm{n}}}{\mathrm{n} !}\)
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