Home > Calculus > Constructing a Maclaurin Series

How do you find the Maclaurin series for #3/(1-2x) #?

Answer 1

The series is #3(1+2x+4x^2+8x^3+16x^4+......)#

Rewriting the expression #3/(1-2x)=3(1-2x)^(-1)# The development of a binomial is #(1+x)^n=1+nx+(n(n-1))/(1.2)x^2+(n(n-1)(n-2))/(1.2.3)x^3+(n(n-1)(n-3)(n-4))/(1.2.3.4)x^2+.....#

So we have #(1-2x)^(-1)=1+2x+(-1(-2))/(1.2)(-2x)^2+(-1(-2)(-3))/(1.2.3)(-2x)^3+(-1(-2)(-3)(-4))/(1.2.3.4)(-2x)^2+.....#

#=1+2x+4x^2+8x^3+16x^4+.....#

and finally #3/(1-2x)=3(1-2x)^(-1)=3(1+2x+4x^2+8x^3+16x^4+....)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Evelyn Agard

To find the Maclaurin series for ( \frac{3}{1 - 2x} ), you can use the geometric series formula:

[ \frac{1}{1 - u} = \sum_{n=0}^{\infty} u^n ]

Substitute ( u = 2x ) into the formula:

[ \frac{1}{1 - 2x} = \sum_{n=0}^{\infty} (2x)^n ]

Multiply both sides by 3:

[ \frac{3}{1 - 2x} = 3 \sum_{n=0}^{\infty} (2x)^n ]

So, the Maclaurin series for ( \frac{3}{1 - 2x} ) is ( 3 \sum_{n=0}^{\infty} (2x)^n ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.