Question

Consider the binomial product (a+b)^n, where the third term of the expansion is 127575x^5y^4 and the fourth term is -354375x^4y^6. Find a, b, and n

Answer 1

Use the binomial Theorem in your book. It has two methods which are both correct; factorial method and Pascal's triangle method

Answer 2

Okay I will try that.

Answer 3

there is another trick to figuring out the coefficients and its pretty cool

example

take \((a+b)^7\)
we know it will look like this

\(a^7 + a^6b+a^5b^2+a^4b^3+a^3b^4+a^2b^5+ab^6+b^7\)

we just need the coefficients in front of each term

the first coefficient is 1, then we do the power of a in the term (7) multiply that by the coefficient in that term (1) and divide by the number term you are in (1)

so we get 7*1/1 = 7
so
\(a^7 + 7a^6b+a^5b^2+a^4b^3+a^3b^4+a^2b^5+ab^6+b^7\)

now multiply the power of the a in that term (6) times the coefficient (7) then divide that by the number term your in (2)
7*6/2 = 21
so

\(a^7 + 7a^6b+21a^5b^2+a^4b^3+a^3b^4+a^2b^5+ab^6+b^7\)\)
5*21/3 = 5*7 = 35
\(a^7 + 7a^6b+21a^5b^2+35a^4b^3+a^3b^4+a^2b^5+ab^6+b^7\)
35*4/4 = 35
\(a^7 + 7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+a^2b^5+ab^6+b^7\)
and now we are done because it will be symmetric.

\(a^7 + 7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7\)

Answer 4

if you practice this for 10 minutes you can compute binomials faster by hand than someone can put them into a calculator....

Answer 5

Aww that is a really cool trick! Thanks for sharing!

Answer 6

fo she z