Binomial expansion (a) Expand (1+e^x)+(e^3x) in ascending powers of x to the term x^2. (b) If the coefficient of x^2 in the expansion ((1+e^(x)-e^(3x))(1-3x)^(n))/(e^(3x) ) is (…

Question

Binomial expansion

(a) Expand (1+e^x)+(e^3x) in ascending powers of x to the term x^2.

(b) If the coefficient of x^2 in the expansion
`((1+e^(x)-e^(3x))(1-3x)^(n))/(e^(3x) )` is
`(433)/(2)` , find n.

Answer 1

Step-by-step explanation: (a) To expand (1 + e^x) + (e^3x) in ascending powers of x to the term x^2, we can use the Taylor series expansion.

The Taylor series expansion for e^x is given by:

e^x = 1 + x + x^2/2! + x^3/3! + ...

So,

e^x = 1 + x + x^2/2! + x^3/3! + ...

and

e^3x = 1 + 3x + 9x^2/2! + 27x^3/3! + ...

Now, we can substitute these expansions into the original equation to get:

(1 + e^x) + (e^3x) = 1 + (1 + x + x^2/2! + x^3/3! + ...) + (1 + 3x + 9x^2/2! + 27x^3/3! + ...)

Expanding, we get:

(1 + e^x) + (e^3x) = 2 + x + 4x + x^2/2! + 9x^2/2! + ...

So, to the term x^2, the expansion becomes:

(1 + e^x) + (e^3x) = 2 + x + 4x + (x^2/2! + 9x^2/2!) + ...

which simplifies to:

(1 + e^x) + (e^3x) = 2 + x + 4x + 11x^2/2! + ...

(b) The coefficient of x^2 in the expansion is 11/2!. So, n = 11/2! = 11/2.