Asked by Samantha
I need help trying to write out these binomial expansions: they all need to be raised to the 8th power. Thanks
1. (x + y) 2. (w + z) 3. (x - y) 4. (2a + 3b)
- Now explain how your answer for #1 could be used as a formula to help you answer each of the other items. In each case, for #2, 3 and 4, tell what would x equal and what would y equal.
- Without writing out the whole polynomial, find the 16th term is (x + y)19. Explain how you constructed it.
1. (x + y) 2. (w + z) 3. (x - y) 4. (2a + 3b)
- Now explain how your answer for #1 could be used as a formula to help you answer each of the other items. In each case, for #2, 3 and 4, tell what would x equal and what would y equal.
- Without writing out the whole polynomial, find the 16th term is (x + y)19. Explain how you constructed it.
Answers
Answered by
Steve
Recall your Pascal's triangle for the expansion of (a+b)^n
On the nth row, you have the kth element: C(n,k)a^(n-k+1)b^(k-1)
That is, on the 8th row you have
C(8,0)a^8b^0 C(8,1)a^7b^1 . . .
= a^8 + 8a^7b + 28a^6b^2 + ... + 28a^2b^6 + 8ab^7 + b^8
There are lots of places online where you can find Pascal's triangle.
So, for this problem, the 16th term of the 19th row is
C(19,15)(2a)^4(3b)^15 = 3876(2a)^4(3b)^15
= 3876(2^5)(3^14)a^4b^15
= 3876(16)(14348907)a^4b^15
= 889861816512a^4b^15
=
On the nth row, you have the kth element: C(n,k)a^(n-k+1)b^(k-1)
That is, on the 8th row you have
C(8,0)a^8b^0 C(8,1)a^7b^1 . . .
= a^8 + 8a^7b + 28a^6b^2 + ... + 28a^2b^6 + 8ab^7 + b^8
There are lots of places online where you can find Pascal's triangle.
So, for this problem, the 16th term of the 19th row is
C(19,15)(2a)^4(3b)^15 = 3876(2a)^4(3b)^15
= 3876(2^5)(3^14)a^4b^15
= 3876(16)(14348907)a^4b^15
= 889861816512a^4b^15
=
Answered by
Steve
Oops. It was just (x+y)^19. My bad.
C(19,15)x^4(y^15 = 3876x^4y^15
C(19,15)x^4(y^15 = 3876x^4y^15
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