Respuesta :
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
Use the generalized binomial expansion to expand (1 + x)^a up to the a = 5 and hence determine (1.05)^5. to 5 decimal places.
Answer:
Using the binomial theorem
[tex](1 .05)^5 = 1.27628[/tex]
Step-by-step explanation:
The binomial theorem is given by
[tex](a +b)^n = \binom{n}{0} a^nb^0 + \binom{n}{1} a^{n-1}b^1....+ \binom{n}{n} a^0b^n[/tex]
For the given case, we have
[tex]a = 1 \\\\b = 0.05 \\\\n = 5 \\\\[/tex]
So,
[tex](1 +0.05)^5 = \binom{5}{0} (1)^5(0.05)^0 + \binom{5}{1} (1)^4(0.05)^1 + \binom{5}{2} (1)^3(0.05)^2 + \binom{5}{3} (1)^2(0.05)^3 + \binom{5}{4} (1)^1(0.05)^4 + \binom{5}{5} (1)^0(0.05)^5[/tex]
[tex](1 +0.05)^5 = (1)(1)^5(0.05)^0 + (5) (1)^4(0.05)^1 + (10) (1)^3(0.05)^2 + (10) (1)^2(0.05)^3 + (5) (1)^1(0.05)^4 + (1) (1)^0(0.05)^5[/tex]
[tex](1 +0.05)^5 = 1 + 0.25 + 0.025 + 0.00125 + 0.00003125 + 0.0000003125[/tex]
[tex](1 + 0.05)^5 = 1.27628[/tex]
Therefore, using the binomial theorem
[tex](1 .05)^5 = 1.27628[/tex]
