Question 1162717: Determine if the statement is true or false.
If the statement is false, then correct it and make it true.
The first term in the expansion of (a + b)^999 is a^999.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the statement is correct.
the first term in the expansion of (a+b)^999 is c(999,0) * a^999 * b^0.
this becomes (a + b)^999 = 1 * a^999 * 1 which becomes a^999.
the general equation is (a+b)^n = the sum of c(n,k) * a^(n-k) * b^k for k = 0 to k = n
note that c(n,k) is equal to n! / (k! * (n-k)!)
an example will help clafify the equation.
consider (a+b)^1
in this case, n = 1.
the formula states that (a+b)^n = the sum of c(n,k) * a^(n-k) * b^k for k = 0 to k = n
k will go from 0 to 1.
therefore, the formula becomes:
(a+b)^1 = c(1,0)*a^1*b^0 + c(1,1)*a^0*b^1
this becomes 1*a^1 + 1*b^1 which becomes a+b.
this makes sense, since it should be intuitively obvious that (a+b)^1 = (a+b) since any quantity raise to the first power is that quantity itself.
consider (a+b)^2
in this case, n = 2.
the formula states that (a+b)^n = the sum of c(n,k) * a^(n-k) * b^k for k = 0 to k = n
k will go from 0 to 2.
therefore, the formula becomes:
(a+b)^2 = c(2,0)*a^2*b^0 + c(2,1)*a^1*b^1 + c(2,2)*a^0*b^2
this becomes 1*a^2 + 2*a*b + 1*b^2 which becomes:
a^2 + 2ab + b^2
this makes sense, since (a+b)^2 is equal to (a+b)*(a+b) which is equal to a*(a+b) + b*(a+b) which is equal to a^2 + ab + ab + b^2 which is equal to a^2 + 2ab + b^2
you can confirm for (a+b)^3 by just following the formula.
in all cases, however, you can see that the first term is a^n
when n is 1, the first term is a^1
then n is 2, the first term is a^2
when n is 999, the first term is a^999
consider (a+b)^999
the first term is c(999,0)*a^999*b^0 which is equal to a^999.
here's some references on the binomial theorem you might find useful.
pick the reference that makes the most sense to you.
don't break your brain.
if you are struggling with the reference, go on to the next one.
they all should be saying the same thing in different ways, some of which may be more understandable in terms of your thinking.
http://hyperphysics.phy-astr.gsu.edu/hbase/alg3.html
https://courses.lumenlearning.com/boundless-algebra/chapter/the-binomial-theorem/
https://people.richland.edu/james/lecture/m116/sequences/binomial.html
https://www.purplemath.com/modules/binomial.htm
https://brilliant.org/wiki/binomial-theorem-n-choose-k/
https://mathbitsnotebook.com/Algebra2/Polynomials/POBinomialTheorem.html
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