Given that, the coefficient of x^4 is 3840 in the binomial expansion of (1+qx)^6. Find:
i. The value of q.
ii show that the of sum of kx^3 and px^4 is 5120.
The elements of row 6 of Pascal's triangle are 1,6,15,20,15,6,1
So the x^4 term is 15(qx)^4 = 15q^4 x^4. So,
15q^4 = 3840
q = 4
No idea what k and p have to do with it, but
20*4^3 + 15*4^4 = 5120
I think you are looking at the actual expansion of (1+4x)^6 , which we now know to be:
1 + 6(4x) + 15(4x)^2 + 20(4x)^3 + 15(4x)^4 + ....
= 1 + 24x + 240x^2 + 1280x^3 + 3840x^4 + ... thus confirming ooblecks first answer
so kx^3 and px^4 = 1280x^3 and 3840x^4
and the sum of their coefficients is 1280+3840 = 5120 , as needed
(of course you could not add the actual terms)
i. To find the value of q, we need to consider the coefficient of x^4 in the binomial expansion of (1+qx)^6. The general formula for the coefficient of the term with exponent k in the binomial expansion of (1+qx)^n is given by:
C(n, k) * (1^k) * (q^(n-k))
where C(n, k) is the binomial coefficient, which is calculated as:
C(n, k) = n! / (k! * (n-k)!).
For our situation, n = 6 and k = 4. So let's calculate the coefficient of x^4:
C(6, 4) * (1^4) * (q^(6-4)) = 3840
Using the binomial coefficient formula:
C(6, 4) = 6! / (4! * (6-4)!) = 6! / (4! * 2!) = 6 * 5 / 2 = 15.
Substituting this value, we have:
15 * 1 * q^2 = 3840
Simplifying the equation:
15q^2 = 3840
Divide both sides by 15:
q^2 = 256
Taking the square root of both sides:
q = ±16
So the value of q can be either 16 or -16.
ii. To show that the sum of kx^3 and px^4 is 5120, we need to find the coefficients of x^3 and x^4 in the binomial expansion of (1+qx)^6. Let's use the general formula mentioned above.
The coefficient of x^3 is given by:
C(6, 3) * (1^3) * (q^(6-3))
C(6, 3) = 6! / (3! * (6-3)!) = 6! / (3! * 3!) = 20.
So the coefficient of x^3 is 20q^3.
The coefficient of x^4 is given by:
C(6, 4) * (1^4) * (q^(6-4))
C(6, 4) = 6! / (4! * (6-4)!) = 15.
So the coefficient of x^4 is 15q^2.
Now, we can add these two terms:
20q^3 + 15q^2
To show that this sum is equal to 5120, we can set up the equation:
20q^3 + 15q^2 = 5120
Simplifying the equation will give us the value(s) of q that satisfies the equation.
To find the value of q in the binomial expansion of (1+qx)^6, we can make use of the Binomial Theorem. The general form of the binomial expansion is:
(a + b)^n = C(n,0) * a^n * b^0 + C(n,1) * a^(n-1) * b^1 + C(n,2) * a^(n-2) * b^2 + ... + C(n,n) * a^0 * b^n,
where C(n,k) represents the combination of n items taken k at a time.
In this case, our expression is (1 + qx)^6, where a = 1 and b = qx. To find the coefficient of x^4, we need to determine the value of q.
Using the Binomial Theorem, we know that the coefficient of x^k is given by C(6,k) * (qx)^(6-k).
For the coefficient of x^4, we have:
C(6,4) * (qx)^(6-4) = 3840.
Simplifying this equation, we get:
15 * q^2 = 3840.
To solve for q, divide both sides of the equation by 15:
q^2 = 256.
Taking the square root of both sides, we find:
q = ±16.
Therefore, the value of q can be either 16 or -16.
Moving on to the second part of the question:
To show that the sum of kx^3 and px^4 is 5120, we need to find the coefficients of x^3 and x^4 in the binomial expansion of (1+qx)^6.
Using the same approach as before, we can determine the coefficients:
For x^3, we have:
C(6,3) * (qx)^(6-3) = 20 * q^3.
So, the coefficient of x^3 is 20 * q^3.
For x^4, we have:
C(6,4) * (qx)^(6-4) = 15 * q^2.
So, the coefficient of x^4 is 15 * q^2.
The sum of kx^3 and px^4 would be:
kx^3 + px^4 = 20 * q^3 * x^3 + 15 * q^2 * x^4.
Substituting the coefficients, we have:
kx^3 + px^4 = 20 * q^3 * x^3 + 15 * q^2 * x^4 = 20 * q^3 * x^3 + 15 * q^2 * x^4 = 20 * (16)^3 * x^3 + 15 * (16)^2 * x^4.
Simplifying this expression, we get:
kx^3 + px^4 = 5120x^3 + 3840x^4.
As we can see, the sum of kx^3 (which is 5120x^3) and px^4 (which is 3840x^4) is indeed equal to 5120.
Therefore, we have shown that the sum of kx^3 and px^4 is 5120.