i need help in this one for the steps.
find the therm that does not contain x in the expansion of ((8x) + ((1)/(2x))^8
Hint: would it occure when the expansion powers were the same (towards the middle?).
Hint #2
Wouldn't the exponent be zero of there is no x term?
To find the term that does not contain x in the expansion of ((8x) + ((1)/(2x))^8, we can use the binomial theorem.
The binomial theorem states that the term of a binomial expansion can be calculated using the formula:
T(n+1) = C(n, r) * a^(n-r) * b^r
where:
- T(n+1) represents the term number you want to find
- C(n, r) represents the binomial coefficient, which can be calculated using nCr, where n is the exponent of the binomial and r is the power of a in the term
- a represents the first term of the binomial, which in this case is 8x
- b represents the second term of the binomial, which in this case is (1/(2x))
- n represents the exponent of the binomial, which in this case is 8
- r represents the power of a in the term, which can range from 0 to n
To find the term that does not contain x, we need to find the term where the power of x is zero (x^0).
So, using the formula, we substitute x^0 for x and simplify:
T(n+1) = C(n, r) * a^(n-r) * b^r
T(n+1) = C(8, r) * (8x)^(8-r) * ((1/(2x)))^r
T(n+1) = C(8, r) * 8^(8-r) * x^(8-r) * (1/(2x))^r
To make sure there is no x term, we set the power of x to zero (8 - r = 0) and solve for r:
8 - r = 0
r = 8
Now, we can substitute r = 8 into the formula and simplify further:
T(n+1) = C(8, 8) * 8^(8-8) * x^(8-8) * (1/(2x))^8
T(n+1) = C(8, 8) * 8^0 * x^0 * (1/(2x))^8
T(n+1) = C(8, 8) * 1 * 1 * (1/(2x))^8
T(n+1) = C(8, 8) * (1/(2x))^8
The term that does not contain x is given by T(9) because n+1 = 8+1 = 9.
Therefore, the term that does not contain x in the expansion of ((8x) + ((1)/(2x))^8 is C(8, 8) * (1/(2x))^8.