Final answer:
a) The expansion of (1+2x^2)^10 as a finite series is 1 + 10(2x^2) + 10(10-1)(2x^2)^2 + ... b) The coefficient of x^6 in the sum is given by 10C6(2x^2)^7. c) The Maclaurin polynomial of degree 4 for f(x)=(1+x)^(5/4) is 1 + (5/4)x + (5/4)(1/2)x^2 + (-3/4)(-1/6)x^3 + (-7/4)(-7/4)(1/24)x^4.
Step-by-step explanation:
a) Express (1+2x2)10 as a finite series:
Using the binomial theorem, we can expand the expression:
(1+2x2)10 = 1 + 10(2x2) + 10(10-1)(2x2)2 + ...
b) Find the coefficient of x6 in the sum:
The general term in the expansion of (1+2x2)10 is given by:
Tr = 10Cr-1(2x2)r
To find the coefficient of x6, we need to find the value of r that satisfies r - 1 = 6, which gives r = 7.
Therefore, the coefficient of x6 in the sum is:
10C6(2x2)7
c) Write out the Maclaurin polynomial of degree 4 for f(x)=(1+x)5/4:
The Maclaurin polynomial is given by:
Pn(x) = f(0) + f'(0)x + f''(0)x2/2! + f'''(0)x3/3! + ... + fn(0)xn/n!
For f(x)=(1+x)5/4, we can calculate the derivatives at x = 0:
f(0) = (1+0)5/4 = 1
f'(x) = (5/4)(1+x)1/4
f''(x) = (5/4)(1/4)(1+x)-3/4
f'''(x) = (-3/4)(-1/4)(1+x)-7/4
f''''(x) = (-7/4)(-7/4)(1+x)-11/4
Substituting these values into the Maclaurin polynomial formula, we get:
P4(x) = 1 + (5/4)x + (5/4)(1/2)x2 + (-3/4)(-1/6)x3 + (-7/4)(-7/4)(1/24)x4