If the expression (x2+2x−1)8 is completely expanded, what is the sum of the coefficients of the terms with even powers of x?
the term is (x^2 + 2x -1)^8 :)
265
i mean 256!!!
To find the sum of the coefficients of the terms with even powers of x in the expansion of the expression (x^2+2x−1)^8, we first need to expand the expression using the binomial theorem.
The binomial theorem states that for any positive integer n, the expansion of (a + b)^n can be calculated using the formula:
(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-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n
In this case, a = x^2, b = 2x, and n = 8.
Now, let's find the sum of the coefficients of the terms with even powers of x.
We can start by expanding the expression using the binomial theorem:
(x^2 + 2x - 1)^8 = C(8, 0) * x^(2*8) * (2x)^0 + C(8, 1) * x^(2*8-1) * (2x)^1 + C(8, 2) * x^(2*8-2) * (2x)^2 + ... + C(8, 7) * x^2 * (2x)^(8-2) + C(8, 8) * x^0 * (2x)^8
This expands to:
x^16 - 16x^15 + 120x^14 - 560x^13 + 1820x^12 - 4368x^11 + 8008x^10 - 11440x^9 + 12870x^8
Now, we need to find the sum of the coefficients of the terms with even powers of x. These terms are x^16, x^14, x^12, x^10, and x^8.
The sum of the coefficients can be found by substituting x = 1 into the expression and adding up the resulting coefficients:
Sum = 1^16 - 16(1^15) + 120(1^14) - 560(1^13) + 1820(1^12) - 4368(1^11) + 8008(1^10) - 11440(1^9) + 12870(1^8)
Simplifying, we get:
Sum = 1 - 16 + 120 - 560 + 1820 - 4368 + 8008 - 11440 + 12870
Sum = 6915
Therefore, the sum of the coefficients of the terms with even powers of x in the expanded expression is 6915.