i) Expand (1-(x/2))1/5
ii)State the range of values of x, for which the expansion is valid
Did you mean
(1 - x/2)^(1/5) ?
Are you studying the binomial theorem?
To expand (1-(x/2))^1/5, we'll use the binomial theorem. The binomial theorem states that for any real number a and any positive integer n, the expansion of (a + b)^n can be found 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
where C(n, k) represents the binomial coefficient, calculated as:
C(n, k) = n! / (k!(n-k)!)
Now, let's apply the above formula to our expression:
(1 - (x/2))^1/5 = C(1/5, 0) * 1^(1/5) * (-x/2)^0
+ C(1/5, 1) * 1^(1/5-1) * (-x/2)^1
+ C(1/5, 2) * 1^(1/5-2) * (-x/2)^2
+ ...
The expansion continues until the last term, which is C(1/5, n) * 1^(1/5-n) * (-x/2)^n.
To find the range of values of x for which the expansion is valid, we need to consider the convergence of the series. The series will converge if the absolute value of the common ratio (x/2) is less than 1.
|x/2| < 1
This inequality can be rearranged to:
-1 < x/2 < 1
Multiplying all parts of the inequality by 2:
-2 < x < 2
Hence, the range of values of x for which the expansion is valid is -2 < x < 2.