If the expression (x^2+2x−1)^8 is completely expanded, what is the sum of the coefficients of the terms with even powers of x?
256
Not sure what's the point of using such a high power. Tedious, but doesn't provide any real insight.
@Steve r u sure of ur answer?
Steve is correct, the sum is 256
I let Wolfram do all the drudgery
http://www.wolframalpha.com/input/?i=expand+%28x%5E2%2B2x−1%29%5E8
Agree with Steve, this question is simply a test of patience.
Thanks, Reiny. I used wolfram also :)
To find the sum of the coefficients of the terms with even powers of x in the expansion of (x^2+2x−1)^8, we need to expand the expression and identify the terms with even powers.
The expansion of a binomial raised to a power can be found using the binomial theorem. According to the binomial theorem, the expansion of (x^2+2x−1)^8 can be written as:
C(8, 0) * (x^2)^8 * (2x)^0 * (-1)^0 + C(8, 1) * (x^2)^7 * (2x)^1 * (-1)^1 + C(8, 2) * (x^2)^6 * (2x)^2 * (-1)^2 + ... + C(8, 8) * (x^2)^0 * (2x)^8 * (-1)^8
where C(n, k) represents the binomial coefficient ("n choose k").
To find the sum of the coefficients of the terms with even powers, we only need to consider terms with even powers of x. In this case, those are the terms where the exponent of x in (x^2)^m * (2x)^n * (-1)^p is an even number.
Let's break down the terms from the expansion:
Term 1: C(8, 0) * (x^2)^8 * (2x)^0 * (-1)^0 (no x terms)
Term 2: C(8, 1) * (x^2)^7 * (2x)^1 * (-1)^1 (odd power of x)
Term 3: C(8, 2) * (x^2)^6 * (2x)^2 * (-1)^2 (even power of x)
And so on...
From the terms above, only Term 3 has an even power of x.
To get the sum of the coefficients of the terms with even powers of x, we add up all the coefficients from those terms.
In this case, we just need to consider the coefficient of Term 3:
C(8, 2) = 28
Therefore, the sum of the coefficients of the terms with even powers of x is 28.