(1+x)^8

To expand the expression (1+x)^8, we can use the binomial theorem which states that for any positive integer n:

(1+x)^n = C(n,0)1^nx^0 + C(n,1)*1^(n-1)*x^1 + C(n,2)*1^(n-2)*x^2 + ... + C(n,n-1)1x^(n-1) + C(n,n)*x^n

where C(n,k) represents the binomial coefficient, calculated as C(n,k) = n! / (k!(n-k)!)

For (1+x)^8, we have:

C(8,0)1^8x^0 + C(8,1)1^7x^1 + C(8,2)1^6x^2 + C(8,3)1^5x^3 + C(8,4)1^4x^4 + C(8,5)1^3x^5 + C(8,6)1^2x^6 + C(8,7)1x^7 + C(8,8)*x^8

Now, let's calculate each term:

C(8,0) = 8! / (0!(8-0)!) = 1 C(8,1) = 8! / (1!(8-1)!) = 8 C(8,2) = 8! / (2!(8-2)!) = 28 C(8,3) = 8! / (3!(8-3)!) = 56 C(8,4) = 8! / (4!(8-4)!) = 70 C(8,5) = 8! / (5!(8-5)!) = 56 C(8,6) = 8! / (6!(8-6)!) = 28 C(8,7) = 8! / (7!(8-7)!) = 8 C(8,8) = 8! / (8!(8-8)!) = 1

Now we substitute these into the expansion:

1x^0 + 81x^1 + 281x^2 + 561x^3 + 701x^4 + 561x^5 + 281x^6 + 81x^7 + 1x^8

Simplifying the terms, we get the expanded form of (1+x)^8 as:

1 + 8x + 28x^2 + 56x^3 + 70x^4 + 56x^5 + 28x^6 + 8x^7 + x^8

So, (1+x)^8 = 1 + 8x + 28x^2 + 56x^3 + 70x^4 + 56x^5 + 28x^6 + 8x^7 + x^8