The first 2 values in ascending power of x in the expansion of (1+x) (2-x/4)^n are p+qx^2.Find the values of n,p,q.
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asked by Raj
today at 12:10pm
(2 - x/4)^n = 2^n - n*2^(n-1) * (x/4) + n(n-1)/2 * 2^(n-2) * (x/4)^2 - ...
= 2^n - n*2^(n-3) *x + n(n-1)*2^(n-7)*x^2 - ...
Now multiply that by 1+x and you have
2^n - n*2^(n-3) *x + n(n-1)*2^(n-7)*x^2 - ... + 2^n*x - n*2^(n-3) *x^2 + n(n-1)*2^(n-7)*x^3 - ...
= 2^n + (2^n - n*2^(n-3))*x + (n(n-1)*2^(n-7) - n*2^(n-3))*x^2 + ...
So, if you want the x term to vanish, you need
2^n - n*2^(n-3) = 0
n = 8
Now that gives you
2^8 + (8(8-1)*2^(8-7) - 8*2^(8-3))*x^2 = 256 - 144x^2 + ...
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👨🏫
oobleck
today at 12:42pm
Why n(n-1)/2 in third term?
An example
(a+b)^5
The binomial expansion gives us
C(5,0)(a^5)(b^0) + C(5,1)(a^4)(b^1) + C(5,2)(a^3)(b^2) + C(5,3)(a^2)(b^3) + C(5,4)(a^1)(b^4) + C(5,5)(a^0)(b^5)
= a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5a b^4 + b^5
look at the coefficients: 1 5 10 10 5 1, they are the 6th row of Pascal's triangle, I hope
that this important property has been taught to you or pointed out to you
recall how we calculate something like C(5,2)
= 5!/(2!3!) = 5*4/2
C(n,2) would be n!/(2!(n-2)!)
= n(n-1)(n-2)(n-3)...(2)(1) / ((n-2)(n-3)...(2)(1) * 2(1) )
= n(n-1)/2!
C(n,4) would be n(n-1)(n-2)(n-3)(n-4)/4!
etc
Well, you see, the third term in the expansion involves n(n-1)/2. It's like when you have a clown juggling balls. The clown has to use both hands to juggle the balls, so you have n balls multiplied by (n-1) balls divided by 2 to account for the fact that the clown is juggling with two hands. It's all about the clown's balancing act! So, in the same way, the n(n-1)/2 term in the expansion is all about juggling and balancing the numbers. Isn't math like a circus sometimes?
In the expansion of (2 - x/4)^n, the general term is given by the binomial theorem as:
C(n, k) * (2^n) * (-x/4)^k
Where C(n, k) is the binomial coefficient or "n choose k" which represents the number of ways to choose k elements from a set of n elements.
In this case, we are interested in the first two terms in ascending powers of x. Let's break down the expansion:
(2 - x/4)^n = 2^n - n*2^(n-1) * (x/4) + (n(n-1)/2) * 2^(n-2) * (x/4)^2 - ...
For the first term, we have (2 - x/4)^0 = 1. So the coefficient of x^0 is 1, which is equal to 2^0.
For the second term, we have (2 - x/4)^1 = 2 - x/4. So the coefficient of x^1 is -1/4, which is equal to -n*2^(n-1).
For the third term, we have (2 - x/4)^2 = 4 - 2x + x^2/16. So the coefficient of x^2 is 1/16, which is equal to (n(n-1)/2) * 2^(n-2).
Since the first two terms are p + qx^2, we can equate the coefficients of x^2 to get:
p = 1/16, q = n(n-1)/2.
Therefore, the values of n, p, and q are n = 8, p = 1/16, and q = 8(8-1)/2 = 28.