use the binomial theorem to expand the binomial
(x+5)^5
I have tried it like 4 times but i cannot get it right
row for n =5 is
1 5 10 10 5 1
1 x^5 5^0
5 x^4 5^1
10 x^3 5^2
10 x^2 5^3
5 x^1 5^4
1 x^0 5^5
so
x^5 + 5 x^4(5) + 10 x^3(25) + 10x^2(125) + 5 x(625) + 3125
which is
x^5 + 25 x^4 + 250 x^3 + 1250 x^2 + 3125 x + 3125
To expand the binomial (x+5)^5 using the binomial theorem, you can use the formula:
(x + y)^n = C(n,0) * x^n * y^0 + C(n,1) * x^(n-1) * y^1 + C(n,2) * x^(n-2) * y^2 + ... + C(n,n) * x^0 * y^n
Where C(n,k) represents the binomial coefficient, defined as:
C(n,k) = n! / (k!(n-k)!)
In our case, (x + 5)^5 becomes:
C(5,0) * x^5 * 5^0 + C(5,1) * x^4 * 5^1 + C(5,2) * x^3 * 5^2 + C(5,3) * x^2 * 5^3 + C(5,4) * x^1 * 5^4 + C(5,5) * x^0 * 5^5
Now, let's calculate each term step by step:
C(5,0) = 5! / (0!(5-0)!) = 1
C(5,1) = 5! / (1!(5-1)!) = 5
C(5,2) = 5! / (2!(5-2)!) = 10
C(5,3) = 5! / (3!(5-3)!) = 10
C(5,4) = 5! / (4!(5-4)!) = 5
C(5,5) = 5! / (5!(5-5)!) = 1
Expanding the binomial:
1 * x^5 * 5^0 + 5 * x^4 * 5^1 + 10 * x^3 * 5^2 + 10 * x^2 * 5^3 + 5 * x^1 * 5^4 + 1 * x^0 * 5^5
Simplifying the terms:
x^5 + 5 * x^4 * 5 + 10 * x^3 * 5^2 + 10 * x^2 * 5^3 + 5 * x^1 * 5^4 + 5^5
Finally, the expanded form of (x+5)^5 is:
x^5 + 25x^4 + 250x^3 + 1250x^2 + 3125x + 3125