Fibonacci?

Question

Expand and simplify using Pascal's Triangle.
(2+3x)^5

Answer 1

Find the 5th row of Pascal's triangle:
........ 1 <-- this is row "0"
........1 1
.......1 2 1
......1 3 3 1
....1 4 .6. 4 1
..1 5 10 10 5 1 <-- this is row "5".

Now you just need to use those as coefficients to the rest of your terms...

Start with 2 and 3x You need all combinations of them where the exponents add to 5:
2^0 * (3x)^5 = 1 * 243x^5 = 243x^5
2^1 * (3x)^4 = 2 * 81x^4 = 162x^4
2^2 * (3x)^3 = 4 * 27x^3 = 108x^3
2^3 * (3x)^2 = 8 * 9x² = 72x²
2^4 * (3x)^1 = 16 * 3x = 48x
2^5 * (3x)^0 = 32 * 1 = 32

Now take each of those terms and multiply them by the values in row 5 of Pascals triangle (1, 5, 10, 10, 5, 1)

243x^5 * 1 = 243x^5
162x^4 * 5 = 810x^4
108x^3 * 10 = 1080x^3
72x² * 10 = 720x²
48x * 5 = 240x
32 * 1 = 32

So the final expansion is:
243x^5 + 810x^4 + 1080x^3 + 720x² + 240x + 32

Answer 2

Don't forget that that expression can be written as

sum k=0 to 5 { 5Ck*2^k*(3x)^(5-k) }

Answer 3

Instead use Newton's Binomial expansion

Answer 4

1(2^5)+
5(2^4(3x))+
10(2^3(3x)^2)+
10(2^2(3x)^3)+
5(2(3x)^4)+
1(3x)^5
Do it from there.

Answer 5

(2+3x)^5
=2^5+5*2^4(3x)+10*2^3(3x)^2+
10*2^2(3x)^3+5*2(3x)^4+(3x)^5
=32+240x+720x^2+1080x^3+
810x^4+243x^5