How do you use the binomial series to expand #(1+2x)^5#?

1 Answer
Jul 12, 2017

Answer:

# (1+2x)^5 = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5 #

Explanation:

Recall the Binomial Theorem

# (a+b)^n = sum_(r=0)^n \ ( (n), (r) ) \ a^(n-r) \ b^r #

Where:

# ( (n), (r) ) =""^nC_r = (n!)/((n-r)!r! #

is the combinatorial, which are also the numbers from the #nth# row of Pascal's Triangle

Thus we have:

# (1+2x)^5 = sum_(r=0)^5 \ ( (5), (r) ) \ (1)^(5-r) \ (2x)^r #
# " " = sum_(r=0)^5 \ ( (5), (r) ) \ (2x)^r #

# " " = ( (5), (0) ) (2x)^(0) + ( (5), (1) ) (2x)^(1) + ( (5), (2) ) (2x)^(2) + ( (5), (3) ) (2x)^(3) + ( (5), (4) ) (2x)^(4) + ( (5), (5)) (2x)^(5) #

# " " = (1) + (5)(2x) + (10) (4x^2) + (10) (8x^3) + (5) (16x^4) + (1) (32x^5) #

# " " = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5 #