Question

Using binomial theorem to determine which number is smaller ${{\left( 1.2 \right)}^{4000}}$ or 800? \[\]

Answer 1

Hint: We recall the expansion of two terms with binomial theorem ${{\left( x+y \right)}^{n}}$ and put $x=1,y=0.2$ to expand ${{\left( 1.2 \right)}^{4000}}={{\left( 1+0.2 \right)}^{4000}}$ binomially. We use the fact that all the terms in a binomial expansion have to be positive if both the binomial terms $x,y$ are positive to check which number is smaller between ${{\left( 1.2 \right)}^{4000}}$ and 800. \[\]

Complete step-by-step solution:
We know that binomial is the algebraic expression involving two terms and each term with a distinct variable. We know that we can use the binomial theorem (or binomial expansion) to describe the algebraic expansion of the power of a binomial. If $x,y$ are the two terms of binomial with some positive integral power $n$ then the binomial expansion is given by;
\[{{\left( x+y \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{y}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{y}^{0}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{y}^{0}}+...+{}^{n}{{C}_{n}}{{x}^{0}}{{y}^{n}}\]
The above expression is called a binomial formula or binomial identity. If the two terms in the binomial are positive then all the terms in the expansion are also positive.
We are asked in the question to determine which number is smaller ${{\left( 1.2 \right)}^{4000}}$ or 800. Let use consider
\[{{\left( 1.2 \right)}^{4000}}={{\left( 1+0.2 \right)}^{4000}}\]
We expand the above terms binomially by using the binomial formula for $x=1,y=0.2,n=4000$ to have;
 \[\begin{align}
  & {{\left( 1+0.2 \right)}^{4000}}={}^{4000}{{C}_{0}}{{1}^{4000}}{{\left( 0.2 \right)}^{0}}+{}^{4000}{{C}_{1}}{{1}^{3999}}{{\left( 0.2 \right)}^{1}}+{}^{4000}{{C}_{2}}{{1}^{3998}}{{\left( 0.2 \right)}^{2}}+...+{}^{4000}{{C}_{2}}{{1}^{0}}{{\left( 0.2 \right)}^{4000}} \\
 & \Rightarrow {{\left( 1+0.2 \right)}^{4000}}=1\times 1\times 1+4000\times 1\times 0.2+t \\
 & \Rightarrow {{\left( 1+0.2 \right)}^{4000}}=1+800+t \\
 & \Rightarrow {{\left( 1+0.2 \right)}^{4000}}=801+t \\
 & \therefore {{\left( 1.2 \right)}^{4000}}=801+t \\
\end{align}\]

Here we have denoted $t={}^{4000}{{C}_{2}}{{1}^{3998}}{{\left( 0.2 \right)}^{2}}+{}^{4000}{{C}_{3}}{{1}^{3997}}{{\left( 0.2 \right)}^{4}}...+{}^{4000}{{C}_{2}}{{1}^{0}}{{\left( 0.2 \right)}^{4000}}$. Since all the terms in the summation expression of $t$ is positive, $t$ will also be positive. So we have
\[\begin{align}
  & \Rightarrow 800<801+t \\
 & \Rightarrow 800<{{\left( 1.2 \right)}^{4000}} \\
\end{align}\]
So 800 is smaller than ${{\left( 1.2 \right)}^{4000}}$.

Note: The combinatorial expressions ${}^{n}{{C}_{k}}\text{ }$for positive integer $k=0,1,2...n$ appearing in the binomial expansion are called binomial coefficients and their sum is ${{2}^{n}}$. We can expand ${{\left( x-y \right)}^{n}}$ by taking${{\left( x+\left( -y \right) \right)}^{n}}$. We can also use binomial expansion to check what remainder a large number ${{z}^{n}}$shall leave when divided by smaller numbers $x,y$ by expanding ${{z}^{n}}={{\left( x+y \right)}^{n}}$ binomially. The binomial expansion for any power $r$is given by, ${{\left( x+y \right)}^{r}}=\sum\limits_{k=0}^{\infty }{{}^{r}{{C}_{k}}}{{x}^{r-k}}{{y}^{k}}$ where ${}^{r}{{C}_{k}}=\dfrac{r\left( r-1 \right)....\left( r-k+1 \right)}{k!}$.