heeeeelllppppp, math problem....please?

Question

find the sum of the numerical coeffeciants in the expansion of (x=y)^10.

THNX!

Answer 1

i guess you mean (x+y)^10.

First the long way (just o make sure what a binomial expansion is and how the coefficients look like).

The expansion is
(10 \choose 0) x^10y^0 +
(10 \choose 1) x^9y^1 +
(10 \choose 2) x^8y^2 +
(10 \choose 3) x^7y^3 +
(10 \choose 4) x^6y^4 +
(10 \choose 5) x^5y^5 +
(10 \choose 6) x^4y^6 +
(10 \choose 7) x^3y^7 +
(10 \choose 8) x^2y^8 +
(10 \choose 9) x^1y^9 +
(10 \choose10) x^0y^10

So the sum of the coefficients is
(10 \choose 0) + (10 \choose 1) + (10 \choose 2) +
(10 \choose 3) + (10 \choose 4) + (10 \choose 5) +
(10 \choose 6) + (10 \choose 7) + (10 \choose 8) +
(10 \choose 9) + (10 \choose10) =
1 + 10 + 45 + 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1
= 1024

We could have used a shortcut as well. In the expansion above, if we had put x = y =1, then each term in the expansion would be exactly the same as its coefficient.
ie,second term would be
(10 \choose 1) 1^9 * 1^1 = (10 \choose 1) = 10.

so if x=y=1 the whole expansion is equal to the sum of the coefficients only.

but if x=y=1, then (x+y)^10 = (1+1)^10 = 2^10 = 1024.

Hope this helps.

Answer 2

You have eleven coefficients, but you need to find only 6, since the first five are repeated again.

Term 1 1
Term 2 10
Term 3 45
Term 4 120
Term 5 210
Term 6 252
So add 252 to twice the sum of the terms 1 thru 5.

Answer 3

I assume you mean (x+y)^10.

Then, use the binomial expansion theorem: And so, you will get:

1+10+45+120+210+252+210+120+45+10+1
1024 is the answer

Answer 4

i assume you mean (x+y)^10
the answer is 2^10 = 1024

whY? because the sum of the coeffs in the expansion is the value of the expansion when x=1 and y=1 which must be the same as (1+1)^10

Answer 5

do you mean (x+y)^10???? If so, we will need to know something else about x or y or how they relate to eachother.

Answer 6

huh?



good luck solving it though!