simplify: (x+y)^10
simplify: (2a+5)^5
I would say expand, not simplify
binomial coefs, use Pascal's triangle (google it if necessary)
for n = 10
1 10 45 120 210 252 then down again
so
x^10 + 10 x^9 y + 45 x^8 y^2 etc
for n = 5
1 5 10 10 5 1
so
(2a)^5 + 5(2a)^4 (5) + 10(2a)^3(5)^2 ...etc
or
32 a^5 + 400 a^4 + 2000 a^3 etc
why thank you!
To simplify the expression (x+y)^10, we can expand it using the binomial theorem. The binomial theorem states that (a+b)^n can be expanded as the sum of the terms of the form C(n, k) * a^(n-k) * b^k, where C(n, k) is the binomial coefficient.
For the expression (x+y)^10, we have:
(x+y)^10 = C(10, 0) * x^(10-0) * y^0 + C(10, 1) * x^(10-1) * y^1 + C(10, 2) * x^(10-2) * y^2 + ... + C(10, 10) * x^(10-10) * y^10
The binomial coefficients can be calculated using the formula C(n, k) = n! / (k! * (n-k)!), where n! denotes the factorial of n.
Let's simplify further:
(x+y)^10 = C(10, 0) * x^10 * y^0 + C(10, 1) * x^9 * y^1 + C(10, 2) * x^8 * y^2 + ... + C(10, 10) * x^0 * y^10
C(10, 0) = 10! / (0! * (10-0)!) = 1
C(10, 1) = 10! / (1! * (10-1)!) = 10
C(10, 2) = 10! / (2! * (10-2)!) = 45
...
C(10, 10) = 10! / (10! * (10-10)!) = 1
Therefore, simplifying (x+y)^10 gives us:
(x+y)^10 = x^10 + 10x^9y + 45x^8y^2 + ...
Similarly, to simplify the expression (2a+5)^5, we follow the same procedure as above:
(2a+5)^5 = C(5, 0) * (2a)^(5-0) * 5^0 + C(5, 1) * (2a)^(5-1) * 5^1 + C(5, 2) * (2a)^(5-2) * 5^2 + ... + C(5, 5) * (2a)^0 * 5^5
Expand and simplify further:
(2a+5)^5 = C(5, 0) * (2a)^5 * 5^0 + C(5, 1) * (2a)^4 * 5^1 + C(5, 2) * (2a)^3 * 5^2 + ... + C(5, 5) * (2a)^0 * 5^5
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, 5) = 5! / (5! * (5-5)!) = 1
Therefore, simplifying (2a+5)^5 gives us:
(2a+5)^5 = (2a)^5 + 5(2a)^4 * 5 + 10(2a)^3 * 5^2 + ... + 5^5