Question:
Find the first 3 terms in the expansion, in ascending powers of x, of
(2 + x)^6. Hence, obtain the coefficient of x^2 in the expansion of
(2 + x - x^2)^6
I know the first part of the question, and the first 3 terms are
(64 + 192x + 240x^2 + ...)
How do i use this answer in the second part??
see other question.
How do you work out the first part of the quarter
*Question
1342,$
Ah, the joys of expansions! Let's take a humorous detour to tackle this one.
So, you have the first three terms of (2 + x)^6, which are (64 + 192x + 240x^2 + ...). Now, you want to find the coefficient of x^2 in the expansion of (2 + x - x^2)^6. It's like expanded algebra is having a wild party and you've just been invited!
To obtain the coefficient of x^2 in the new expansion, you'll need to take a deep breath and carefully analyze your original answer. You know, it's important to handle coefficients with care – they're sensitive little creatures.
Since the original answer shows the coefficients of x^0, x^1, x^2, and so on, you can easily find the coefficient of x^2. In this case, it's 240 (from the 240x^2 term). Just like that, you've plucked it from the expansion garden!
I hope this humor-laden explanation helped you understand how to use the information from the first part of your question to find the coefficient of x^2 in the second part. Now, go forth and conquer those expansions!
In order to obtain the coefficient of x^2 in the expansion of (2 + x - x^2)^6, we need to find the coefficient of x^2 in each of the terms (64 + 192x + 240x^2 + ...) and then add them up.
Let's break down the terms in the expansion (2 + x - x^2)^6:
(2 + x - x^2)^6 = (2 + (x - x^2))^6
We can use the binomial theorem to expand this expression. According to the binomial theorem, the coefficient of x^k in the expansion of (a + b)^n can be calculated using the formula: C(n, k) * a^(n-k) * b^k, where C(n, k) is the binomial coefficient (also known as "n choose k").
Now, let's use this formula to find the coefficient of x^2 in each term of the expansion. We already know the first three terms of the expansion are: 64 + 192x + 240x^2 + ...
For the first term, 64 = C(6, 0) * 2^(6-0) * (x - x^2)^0 (since x^0 = 1 for any x)
So, the coefficient of x^2 in the first term is 0.
For the second term, 192x = C(6, 1) * 2^(6-1) * (x - x^2)^1
The coefficient of x^2 in the second term is also 0, since (x - x^2)^1 involves only x^1 and x^0 terms.
For the third term, 240x^2 = C(6, 2) * 2^(6-2) * (x - x^2)^2
Here, we can see that (x - x^2)^2 will have terms involving x^2. Using the formula for the binomial coefficients, we can calculate the coefficient as follows:
C(6, 2) = 6! / (2! * (6-2)!) = 15
2^(6-2) = 2^4 = 16
Thus, the coefficient of x^2 in the third term is 15 * 16 = 240.
Now, to obtain the coefficient of x^2 in the entire expansion, we add up the coefficients of x^2 from each of the terms: 0 + 0 + 240 + ...
Since we know the first three terms, and the fourth and subsequent terms continue with higher powers of x, we can safely assume that the coefficient of x^2 in the expansion of (2 + x - x^2)^6 is 240.