expand using the binomial theorem: (2x-1)^5?

Answer 1

love it

Answer 2

Binomial Theorum:

(x+y)^n = Σ(n k) (x^k)y^(n-k)

(2x -1)^5 = Σ (5 k) (2x^k) (-1^ (5-k))

= (2x)^5 + 5(2x)^4(-1) + 10(2x)^3(-1)^2 + 10(2x)^2(-1)^3 + 5(2x)(-1)^4 + (-1)^5)
since -1^n = -1 when n is odd
= (2^5)*x^5 - 5(2^4)x^4 + 10(2^3)x^3 - 10(2^2)x^2 + 5(2^1)x - 1
= 32x^5 -80x^4 + 80x^3 - 40x^2 + 20x -1

Answer 3

(a + b) ^ n = a ^ 5 + 5a ^ 4 * b + 10 a ^ 3 * b ^ 2 + 10a ^ 2 * b ^ 3 + 5ab ^ 4 + b ^ 5

then substitute a = 2x, b = -1

Answer 4

to amplify the formula (a+b)^n, you persist with pascal's triangle built as follows: ........ a million................... for n=0 ....... a million a million................ for n=a million .......a million 2 a million ................for n=2 ......a million 3 3 a million..............for n=3 .....a million 4 6 4 a million............for n=4 ...a million 5 10 10 5 a million......for n=5 .a million 6 15 20 15 6 a million...for n=6 each and each decision in the triangle is the sum of both promptly above it. those numbers many times is the coefficients of the better formula. as for the variables, you start up with the optimal degree (n) for a and the bottom degree (0) for b and progressively advance by one for a and shrink by a million for b; so the first portion of the expression will be a^6b^0 (meaning a^6) and the only ideal one will be a^0b^6 (meaning b^6). as for the signal, if the formula is (a+b)^n, the signal is continually effective (+). for (a-b)^n, you commerce between + and -, starting up with +. so, the only ideal effect will be: (a-b)^6 = a^6 - 6a^5b + 15a^4b^2 - 20a^3b^3 + 15a^2b^4 - 6ab^5 + b^6 for (2x-5)6, replace a with 2x and b with 5. wish that solutions your question!!

Answer 5

I get 32x^5 - 80x^4 + 80x^3 - 40x^2 + 10x - 1

Answer 6

(2x - 1)^5 = 2^5x^5 + (-1)5*2^4x^4 + (-1)^2(5*4/2)2^3x^3 + (-1)^3((5*4*3/(1*2*3))2^2x^2 + (-1)^4(5)(2)x + (-1)^5

Answer 7

for binomial theorem there is formula
Tr+1=nCr(x)^(n-r) y^r
if form is (x+y)^n
use it to find ur ans