Advertisements
Advertisements
Question
Find the coefficient of x4 in the expansion `(3 - 4x + x^2)/"e"^(2x)`
Advertisements
Solution
`(3 - 4x + x^2)/"e"^(2x) = (3 - 4x + x^2) "e"^(-2x)`
= `(3 -4x + x^2) [1 + (-2x)/(1!) + (-2x)^2/(∠2) + (-2x)^3/(∠3) ...]`
Coeffiient of x4: `3[(-2)^4/(4!)] - 4[(-2)^3/(3!)] + 1[(-2)^2/(2!)]`
= `3[16/24] + (- 4) ((- 8))/6 + 4/2`
= `48/24 + 32/6 + 2`
= `2 + 16/3 + 2`
= `(6 + 16 + 6)/3`
`28/3`
APPEARS IN
RELATED QUESTIONS
Expand the following in ascending powers of x and find the condition on x for which the binomial expansion is valid
`1/(5 + x)`
Find `root(3)(10001)` approximately (two decimal places
Prove that `root(3)(x^3 + 6) - root(3)(x^3 + 3)` is approximately equal to `1/x^2` when x is sufficiently large
Prove that `sqrt((1 - x)/(1 + x))` is approximately euqal to `1 - x + x^2/2` when x is very small
Write the first 6 terms of the exponential series
e5x
Write the first 4 terms of the logarithmic series
log(1 – 2x) Find the intervals on which the expansions are valid.
Write the first 4 terms of the logarithmic series
`log((1 + 3x)/(1 -3x))` Find the intervals on which the expansions are valid.
Find the value of `sum_("n" = 1)^oo 1/(2"n" - 1) (1/(9^("n" - 1)) + 1/(9^(2"n"- 1)))`
Choose the correct alternative:
The coefficient of x6 in (2 + 2x)10 is
Choose the correct alternative:
If (1 + x2)2 (1 + x)n = a0 + a1x + a2x2 + …. + xn + 4 and if a0, a1, a2 are in AP, then n is
Choose the correct alternative:
The sum up to n terms of the series `1/(sqrt(1) +sqrt(3)) + 1/(sqrt(3) + sqrt(5)) + 1/(sqrt(5) + sqrt(7)) + ...` is
Choose the correct alternative:
The sum of an infinite GP is 18. If the first term is 6, the common ratio is
Choose the correct alternative:
The coefficient of x5 in the series e-2x is
Choose the correct alternative:
The value of `1/(2!) + 1/(4!) + 1/(6!) + ...` is
Choose the correct alternative:
The value of `1 - 1/2(2/3) + 1/3(2/3)^2 1/4(2/3)^3 + ...` is
