In Exercise 24, write out the first 5 terms of the Binomial series with the given k-value.
k = 4
The Binomial series formula is:
\((1 + x)^b =\sum_{n=0}^∞\)\(b\choose n\)\(x^n\)
So the expansion of the first 5 terms is:
\((1 + k)^5 = \frac{1}{0!}x^0 +
\frac{k}{1!}x^1 + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 +
\frac{k(k-1)(k-2)(k-3)}{4!}x^4\) \((1 +
k)^5 = 1 + kx + \frac{k(k-1)}{2!}x^2 + \frac{k(k-1)(k-2)}{3!}x^3 +
\frac{k(k-1)(k-2)(k-3)}{4!}x^4\)
Then we can plug in k = 4:
\((1 + k)^5 = 1 + 4x + \frac{4(4-1)}{2!}x^2 +
\frac{4(4-1)(4-2)}{3!}x^3 +
\frac{4(4-1)(4-2)(4-3)}{4!}x^4\)
\((1 + k)^5 = 1 + 4x + \frac{4(3)}{2}x^2 +
\frac{4(3)(2)}{6}x^3 + \frac{4(3)(2)(1)}{24}x^4\)
\((1 + k)^5 = 1 + 4x + \frac{12}{2}x^2 +
\frac{24}{6}x^3 + \frac{24}{24}x^4\)
\((1 + k)^5 = 1 + 4x + 6x^2 + 4x^3 +
1x^4\)
When reduced, the first 5 terms of of the Binomial series with k =
4:
\(= 1 + 4x + 6x^2 + 4x^3 + x^4\)