Probably easy, but im stuck.

[cole92]

ok...i got a problem today that i need help on. here is the problem given:

"Find the sum of the numerical coefficients of (x+y)^n. (HINT: Find the sum of the coefficiants of (x+y)^1, (x+y)^2, and so on)."

thanks in advance for anybody that can help me...i really appreciate it..

thanks again

 

[pka]

\(\displaystyle \L
\begin{array}{l}
\left( {x + y} \right)^1 = x + y\quad \Rightarrow \quad Sum = 2 \\
\left( {x + y} \right)^2 = x^2 + 2xy + y^2 \quad \Rightarrow \quad Sum = 4 \\
\left( {x + y} \right)^3 = x^3 + 3x^2 y + 3xy^2 + y^3 \quad \Rightarrow \quad Sum = 8 \\
\end{array}\)

Do you recognize 2,4,8,…? See if you can generalize!

 

[cole92]

great..thanks for the help, but theres one problem...

i understand how to get the x^2+2xy+y^2 and stuff, but where does the sum = 2, sum=4, sum=8, etc come from? and what exactly would be the answer?

thanks again for the help guys

 

[pka]

Look, this is YOUR problem.
YOU need to find the solution.
I have known the solution for years!
But the point is for YOU TO FIND IT.
There is no free lunch unless you can get Soroban involved.

 

[steve_b]

To answer your one question:

in the expansion x^3 + 3x^2 y + 3 xy^2 + y^3

add up the coefficients of each term: 1 [from the x^3 term] + 3 [from x^2 y] + 3 [from xy^2] + 1 [from y^3]

The sum = 1 + 3 + 3 + 1 = 8.

Steve

 

[cole92]

you could have been more elaborate and explained a little.....i got my answer as:

2^n........................can you at least tell me if im right?

 

[tkhunny]

The idea would be for you to prove it.

If the coefficients of \(\displaystyle (x+y)^{n}\) sum to \(\displaystyle 2^{n}\), what can you say about the coefficients of \(\displaystyle (x+y)^{n}*(x+y)\)?

Once you have proven it, there is no more need for clarification.

 

[TchrWill]

i got a problem today that i need help on. here is the problem given:

"Find the sum of the numerical coefficients of (x+y)^n. (HINT: Find the sum of the coefficiants of (x+y)^1, (x+y)^2, and so on)."

thanks in advance for anybody that can help me...i really appreciate it..

thanks again

Take a look at the binomial expansion of (a + b)^n. Lets axpand a few with increasing values of n.

(a + b)^0 = 1
(a + b)^1 = 1a + 1b
(a + b)^2 = 1a^2 + 2ab + 1b^2
(a + b)^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^2
(a + b)^4 = 1a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + 1b^4

By now you must have recognized that the coefficients of the terms in each expansion are the values from Pascal's triangle. How elegant. We can write the expansion of (a + b)^n for any "n" by simply referring to Pascal's triangle. The coefficients of (a + b)^n are the numbers from the nth row of Pascals triangle.

Pascal's Triangle
Row.....................................................................................................Sum
0...........................................................................1...............................1
1.......................................................................1......1...........................2
2....................................................................1.....2.....1........................4
3.................................................................1....3......3.....1....................8
1..............................................................1....4.....6......4.....1...............16
5...........................................................1....5...10....10.....5.....1............32
6........................................................1....6...15....20....15....6.....1.........64
7.....................................................1....7...21...35...35....21.....7.....1....etc.
8.................................................1....8...28...56...70....56...28....8.....1
9..............................................1....9...36..84..126..126..84....36.....9.....1
10........................................1...10..45.120..210.252..210..120...45...10.....1

The sum of these coefficients should be now be obvious.
..

 

[Denis]

you could have been more elaborate and explained a little.....i got my answer as:
2^n........................can you at least tell me if im right?

Yes. But whoopseedoo: did you want fries with that?

Before posting your problem, did you bother to:
look up what a coefficient is?
expand a few expressions by yourself, like (x + y)^2 = x^2 +2xy + y^2 ?