Find the fifth term in the expansion of (4-1/5x)^9 in ascending powers of x

You must know the binomial expansion and the general term of the expansion!!

5th term = C(9,4) (4^4)(-x/5)^5
= 126 (256)(-x^5)/3125
= -32256/3125 x^5

oops, just did in descending powers, so

C(9,5) 4^5 (-x/5)^4
= 126(1024)(x^4/625)
= 129024/625 x^4 , in ascending order

To find the fifth term in the expansion of (4-1/5x)^9 in ascending powers of x, we can use the binomial theorem. The binomial theorem states that for any real number a and b, and any positive integer n:

(a + b)^n = C(n, 0)a^n + C(n, 1)a^(n-1)b + C(n, 2)a^(n-2)b^2 + ... + C(n, n-1)ab^(n-1) + C(n, n)b^n

Where C(n, r) represents the binomial coefficient, which is given by the formula:

C(n, r) = n! / (r!(n-r)!)

In our case, a = 4 and b = -1/5x. So, we need to expand (4-1/5x)^9 using the binomial theorem, and then identify the coefficient of the fifth term.

First, let's write out the expansion using the binomial theorem:

(4-1/5x)^9 = C(9, 0)4^9 - C(9, 1)4^8(1/5x) + C(9, 2)4^7(1/5x)^2 - C(9, 3)4^6(1/5x)^3 + C(9, 4)4^5(1/5x)^4 - C(9, 5)4^4(1/5x)^5 + ...

We can see that the fifth term is represented by C(9, 4)4^5(1/5x)^4. So, let's calculate it:

C(9, 4) = 9! / (4!(9-4)!) = 9! / (4!5!) = (9 * 8 * 7 * 6) / (4 * 3 * 2 * 1) = 126

The fifth term is then:

126 * 4^5 * (1/5x)^4 = 126 * 1024 * (1/5)^4 * x^-4

Simplifying further:

= 126 * 1024 * 1/625 * 1/x^4

= 1306368/x^4

Therefore, the fifth term in the expansion of (4-1/5x)^9 in ascending powers of x is 1306368/x^4.