Find the coefficient of x^4y^3 in the expansion of (x+2y)^7
Why did the scarecrow win an award? Because he was outstanding in his field!
To find the coefficient of x^4y^3 in the expansion of (x + 2y)^7, we need to use the binomial theorem. According to the theorem, the coefficient can be found by using the formula:
C(n, r) * a^(n-r) * b^r,
where C(n, r) is the binomial coefficient, a is the coefficient of the first term (x in this case), b is the coefficient of the second term (2y), n is the power to which the binomial is raised (7), and r is the power of the second term that we're interested in finding the coefficient for (3 in this case).
Plugging in the values, we get:
C(7, 3) * (x)^(7-3) * (2y)^3 = 35 * (x^4) * (8y^3) = 280x^4y^3.
So, the coefficient of x^4y^3 in the expansion of (x + 2y)^7 is 280.
To find the coefficient of a term in a binomial expansion, we can use the binomial theorem. According to the theorem, the coefficient of the term x^r * y^n, where r is the power of x and n is the power of y, is given by:
Coefficients = (nCr) * (a^r) * (b^n-r)
Where nCr represents the binomial coefficient, a represents the coefficient of x, and b represents the coefficient of y.
In this case, we need to find the coefficient of the term x^4 * y^3 in the expansion of (x+2y)^7.
Using the binomial theorem, we find:
Coefficients = (7C3) * (x^4) * (2y)^3
Let's calculate the values:
(7C3) = 7! / (3! * (7-3)!)
= 7! / (3! * 4!)
= (7 * 6 * 5) / (3 * 2 * 1)
= 35
(x^4) = x^4
(2y)^3 = 2^3 * (y^3)
= 8y^3
Now, let's substitute these values back into the formula:
Coefficients = 35 * (x^4) * (8y^3)
= 280 * x^4 * y^3
Therefore, the coefficient of x^4y^3 in the expansion of (x+2y)^7 is 280.
To find the coefficient of a specific term in a binomial expansion, we can use the Binomial Theorem.
The Binomial Theorem states that for a binomial expression (a + b)^n, the expansion can be written as the sum of terms of the form (n choose k) * a^(n-k) * b^k, where (n choose k) represents the binomial coefficient.
In this case, the expression is (x + 2y)^7, and we want to find the coefficient of the term with x^4y^3.
Using the formula mentioned above, the coefficient of x^4y^3 will be determined by the binomial coefficient (7 choose 3). This can be calculated using the formula:
(7 choose 3) = 7! / (3! * (7-3)!) = 7! / (3! * 4!)
Next, we substitute the values for the factorials:
7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040
3! = 3 * 2 * 1 = 6
4! = 4 * 3 * 2 * 1 = 24
Plugging in these values:
(7 choose 3) = 5040 / (6 * 24) = 5040 / 144 = 35
Therefore, the coefficient of x^4y^3 in the expansion of (x + 2y)^7 is 35.
Review of binomial expansion:
http://www.purplemath.com/modules/binomial.htm
(a+b)^n
=a^n+na^(n-1)b+(n(n-1)/2)a^(n-2)b²+...+nab^(n-1)+b^n
=C(n,0)a^n+C(n,1)a^(n-1)b+C(n-2)a^(n-2)b^sup2;.....C(n,n)b^n
For n=1
(a+b)1=a+b
(a+b)2=a²+2ab+b^sup2;
(a+b)3=a³+3a^sup2;b+3ab²+b³
...
The coefficients are therefore
1,1
1,2,1
1,3,3,1
1,4,6,4,1
1,5,10,10,5,1
1,6,15,20,15,6,1
1,7,21,35,35,21,7,1
...
This is the Pascal's triangle
http://www.mathsisfun.com/pascals-triangle.html
the sum of each pair of adjacent numbers gives rise to a number on the next line.
This also means that
(a+b)7
=a^7+7a^6b+21a^5b²+35a^4b³+....+7ab^6+b^7 ............(A)
which is almost the solution to the problem.
To solve the problem using the Pascal's triangle, substitute a=x, and b=2y in equation (A) to give the expression of the final answer.
I strongly suggest you review the subject on binomial expansion and Pascal's triangle before proceeding to solve the problem. If you do not have a textbook, use the links given above.