Binomial expansion is used to factor expressions that can be expressed as the power of the sum of two numbers.
The proof that (1.01)^12 exceeds (1.02)^6 by 0.0007 is[tex]\mathbf{(1.01)^{12} - (1.02)^6 \approx 0.0007 }[/tex]
The expressions are given as:
[tex]\mathbf{(1.01)^{12}\ and\ (1.02)^6}[/tex]
A binomial expression is represented as:
[tex]\mathbf{(a + b)^n = \sum\limits^n_{k=0}^nC_k a^{n - k}b^k}[/tex]
Express 1.01 as 1 + 0.01
So, we have:
[tex]\mathbf{(1.01)^{12} = (1 + 0.01)^{12}}[/tex]
Apply the above formula
[tex]\mathbf{(1.01)^{12} = ^{12}C_0 \times 1^{12 - 0} \times 0.01^0 + ......... .......... + ^{12}C_{12} \times 1^{12 - 12} \times 0.01^{12} }}[/tex]
[tex]\mathbf{(1.01)^{12} = 1 \times 1 \times 1 + ......... .......... + 1 \times 1 \times 10^{-24} }}[/tex]
[tex]\mathbf{(1.01)^{12} = 1 + ......... .......... + 10^{-24} }}[/tex]
This gives
[tex]\mathbf{(1.01)^{12} = 1.1268\ (approximated)}[/tex]
Similarly,
Express 1.02 as 1 + 0.02
So, we have:
[tex]\mathbf{(1.02)^6 = (1 + 0.02)^6}[/tex]
Apply [tex]\mathbf{(a + b)^n = \sum\limits^n_{k=0}^nC_k a^{n - k}b^k}[/tex]
[tex]\mathbf{(1.02)^6 = ^6C_0 \times 1^{6 - 0} \times 0.02^0 + ^6C_1 \times 1^{6 - 1} \times 0.02^1 +.............. + ^6C_6 \times 1^{6 - 6} \times 0.02^6 }[/tex][tex]\mathbf{(1.02)^6 = 1 \times 1 \times 1 + 6 \times 1 \times 0.02 +.............. + 1 \times 1 \times 6.4 \times 10^{-11} }[/tex]
[tex]\mathbf{(1.02)^6 = 1 + 0.12 +.............. + 6.4 \times 10^{-11} }[/tex]
This gives
[tex]\mathbf{(1.02)^6 = 1.1261\ (approximated) }[/tex]
Calculate the difference as follows:
[tex]\mathbf{(1.01)^{12} - (1.02)^6 \approx 1.1268 - 1.1261 }[/tex]
[tex]\mathbf{(1.01)^{12} - (1.02)^6 \approx 0.0007 }[/tex]
The above equation means that:
(1.01)^12 exceed the value of (1.02)^6 by 0.0007
Read more about binomial expansions at:
https://brainly.com/question/9554282
