the Q is :
Given that (1+ax)^n = 1 + 30x +375x^2+.... find the values of the constants a and n .
I know that :
(i am writing (n 1) as the coefficient of n)
(n 1) *ax = 30x
(n 2) *(ax)^2 = 375x^2
which can be written as
(n 1)* a = 30
(n 2)*a^2 = 375
I solved this only by intuition and got n = 6 and a = 5
but does anyone know how to calculate it otherwise?
Remember Pascal's triangle?
1
11
121 n=2
1331
14641 n=4
1,5,10,10,5,1
1,6,15,20,15,6,1 n=6
your coefficents are 1,30,375
a*something=30
a^2*next coefficent=375
So, your a=6, then n=6 (last row) and verify by a^2*15=375 yes, you are correct.
Or
The first 3 terms of (1 + ax)^n
= 1 + n(ax) + n(n-1)/2(a^2x^2) + ...
so nax = 30x
na = 30
a = 30/n
and
n(n-1)/2(a^2x^2) = 375x^2
n(n-1)(900/n^2) = 750
900(n-1) - 750n
n = 5
then a = 30/5 = 6
Well, solving equations can sometimes be quite the puzzle, but I'll give it my best shot! Let's see if we can figure out the values of a and n in a more methodical way.
Starting with equation (n 1) * ax = 30x, we can divide both sides of the equation by x to get (n 1) * a = 30.
Similarly, for equation (n 2) * (ax)^2 = 375x^2, let's divide both sides by x^2 to simplify it. This gives us (n 2) * a^2 = 375.
Now, we've got a system of two equations:
(n 1) * a = 30
(n 2) * a^2 = 375
To solve this, we can rearrange the first equation to express n 1 in terms of a:
n 1 = 30/a
Substituting this value into the second equation:
(30/a) * a^2 = 375
Simplifying further, we have:
30a = 375
Dividing both sides by 30:
a = 375/30 = 12.5
Now, substituting the value of a back into the first equation:
(n 1) * 12.5 = 30
Simplifying:
(n 1) = 30/12.5 = 2.4
Since n should be a whole number, it seems like we made a calculation error somewhere. But hey, even math can be a bit of a clown sometimes! Feel free to double-check the calculations and see if you get different results.
To solve for the values of a and n, we can use the equation (1+ax)^n = 1 + 30x + 375x^2 + ...
Step 1: Expand the left side of the equation using the binomial theorem:
(1+ax)^n = 1 + n(ax) + (n(n-1)/2)(ax)^2 + ...
Step 2: Compare the coefficients of the expanded terms with the given equation coefficients:
The coefficient of x in the expanded equation is n*a.
Comparing it to the coefficient of x in the given equation (30x), we have:
n*a = 30 -- equation (1)
The coefficient of x^2 in the expanded equation is (n(n-1)/2)*a^2.
Comparing it to the coefficient of x^2 in the given equation (375x^2), we have:
(n(n-1)/2)*a^2 = 375 -- equation (2)
Step 3: Solve the system of equations (1) and (2) to find the values of a and n.
From equation (1), we can solve for a:
a = 30/n
Substituting this value of a in equation (2), we get:
(n(n-1)/2)*((30/n)^2) = 375
Simplifying, we have:
(n(n-1)/2)*(900/n^2) = 375
(n(n-1)*900) = 750n^2
900n^2 - 900n - 750n^2 = 0
150n^2 - 900n = 0
150n(n-6) = 0
From this, we can see that either n = 0 or n - 6 = 0. However, setting n = 0 does not make sense in the context of the binomial theorem, so we can ignore that solution. Hence, n -6 = 0, which gives us n = 6.
Substituting n = 6 back into equation (1), we can solve for a:
a = 30/n
a = 30/6
a = 5
Therefore, the values of a and n that satisfy the given equation are a = 5 and n = 6.
To find the values of the constants a and n in the equation (1+ax)^n = 1 + 30x + 375x^2 + ..., you can use the coefficients of the terms on the right-hand side of the equation.
Using the properties of binomial expansions, the coefficient of the k-th term in the expansion of (1+ax)^n is given by the binomial coefficient (n k) multiplied by a^k.
From the equation, we can see that the coefficient of the first term (1) is given by (n 0) * a^0, which is equal to 1.
The coefficient of the second term (30x) is given by (n 1) * a^1, which is equal to 30.
The coefficient of the third term (375x^2) is given by (n 2) * a^2, which is equal to 375.
So, we have the following equations:
(n 0) * a^0 = 1 -- Equation (1)
(n 1) * a^1 = 30 -- Equation (2)
(n 2) * a^2 = 375 -- Equation (3)
To find the values of a and n, we can solve these equations simultaneously. Here's one way to approach it:
From Equation (1), we know that a^0 = 1, so a = 1.
Now, Equation (2) simplifies to (n 1) = 30.
Substituting the value of a in Equation (3), we get (n 2) = 375.
Now, we can use the binomial coefficient formula to calculate these values.
(n 1) represents the number of ways to choose 1 element from n, which is equal to n.
Therefore, n = (n 1) = 30.
Similarly, (n 2) represents the number of ways to choose 2 elements from n, which is equal to n(n - 1)/2.
Substituting n = 30 in the formula, we get (n 2) = 30 * 29 / 2 = 435.
So, (n 2) = 435.
Since we already found that a = 1, we have a = 1 and n = 30.
Therefore, the values of the constants a and n are a = 1 and n = 30.
You don't need to rely on intuition alone to solve this problem; you can use the properties of binomial coefficients to calculate the values of a and n analytically.