Question

Which is larger \[{\left( {1.01} \right)^{1000000}}\]or 10,000? Find the $17^{th}$ and $18^{th}$ terms in the expansion\[{\left( {2 + a} \right)^{50}}\]are equal.

Answer 1

Hint: To solve this question we must have the idea about the Binomial expansion of\[{\left( {1 + x} \right)^n}\]. First we have to express 1.01 as the sum of 1 and 0.01. Then we have to expand \[{\left( {1.01} \right)^{1000000}}\] using the Binomial theorem to evaluate it. Then the obtained must be compared to 10,000 and determine which is larger.

Complete step by step answer:
We know that an algebraic expression containing two terms is called a binomial expression. The Binomial theorem states that
“For any positive integer n, the nth power of the sum of two numbers y and x may be expressed as the sum of \[n+1\] terms of the form which are given by
\[{{\left( y+x \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}{{y}^{n-r}}{{x}^{r}}}\]
Or, \[{{\left( y+x \right)}^{n}}={}^{n}{{C}_{0}}{{y}^{n}}+{}^{n}{{C}_{1}}{{y}^{n-1}}x+{}^{n}{{C}_{2}}{{y}^{n-2}}{{x}^{2}}+.............+{}^{n}{{C}_{n-1}}y{{x}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{n}}\]….. (1)
Where,
\[{}^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}\]………………… (2)
Now for the expansion of substituting\[y=1\]in eq. (1) we will get
\[{{\left( 1+x \right)}^{n}}={}^{n}{{C}_{0}}{{1}^{n}}+{}^{n}{{C}_{1}}{{1}^{n-1}}x+{}^{n}{{C}_{2}}{{1}^{n-2}}{{x}^{2}}+..........+{{x}^{n}}\]
\[{{(1+x)}^{n}}=1+nx+\dfrac{n(n-1)}{2!}{{x}^{2}}+...........+{{x}^{n}}\]………………….. (3)
We have to find out\[{{\left( 1.01 \right)}^{1000000}}\]. Let’s express 1.01 as\[\left( 1+0.01 \right)\]. Now considering the expression \[{{\left( 1+0.01 \right)}^{1000000}}\]as \[{{\left( 1+x \right)}^{n}}\]let’s substitute \[x=0.01\]and \[n=1000000\]in the eq. (3) we will get
\[{{\left( 1+0.01 \right)}^{1000000}}=1+1000000\times 0.01+..........\]
Or, \[{{\left( 1+0.01 \right)}^{1000000}}=10,001+.........\]
Here we see that in the expansion 10,001 is added to the positive terms. So the value to be obtained will be greater than 10,000.
Therefore \[{{\left( 1.01 \right)}^{1000000}}\]is larger than 10,000.
Now substituting \[b=2\]and\[n=50\]in eq. (1) we will get the expansion for \[{{\left( 2+a \right)}^{50}}\]which is given by,
\[{{\left( 2+a \right)}^{50}}=\sum\limits_{r=0}^{50}{{}^{50}{{C}_{r}}{{2}^{50-r}}{{a}^{r}}}\]
\[{{\left( 2+a \right)}^{50}}={}^{50}{{C}_{0}}{{2}^{50}}+{}^{50}{{C}_{1}}{{2}^{50-1}}{{a}^{1}}+{}^{n}{{C}_{2}}{{2}^{50-2}}{{a}^{2}}+.............+{}^{50}{{C}_{50-1}}2{{a}^{50-1}}+{}^{50}{{C}_{50}}{{b}^{50}}\]……………. (4)
From the Binomial theorem we see that the Binomial expansion of \[{{\left( 2+a \right)}^{50}}\]contains \[50+1=51\]terms. From eq. (4), we get that
The $1^{st}$ term of \[{{\left( 2+a \right)}^{50}}\]is \[{}^{50}{{C}_{0}}{{2}^{50}}={}^{50}{{C}_{1-1}}{{2}^{50-(1-1)}}{{a}^{1-1}}\]
The $2^{nd}$ term of \[{{\left( 2+a \right)}^{50}}\]is \[{}^{50}{{C}_{1}}{{2}^{50-1}}{{a}^{1}}={}^{50}{{C}_{2-1}}{{2}^{50-(2-1)}}{{a}^{2-1}}\]
The $3^{rd}$ term is \[{}^{50}{{C}_{2}}{{2}^{50-2}}{{a}^{2}}={}^{50}{{C}_{3-1}}{{2}^{50-(3-1)}}{{a}^{3-1}}\]
And so on
Similarly \[{{r}^{th}}\] term is given by \[{}^{50}{{C}_{r-1}}{{2}^{50-(r-1)}}{{a}^{r-1}}\]
Now we will find the ${{17}^{th}}\text{ and }{{18}^{th}}$ term in the expansion \[{{\left( 2+a \right)}^{50}}\]which is given by
\[{{T}_{17}}={}^{50}{{C}_{17-1}}{{2}^{50-(17-1)}}{{a}^{17-1}}={}^{50}{{C}_{16}}{{2}^{34}}{{a}^{16}}\]………………………………… (5)
And \[{{T}_{18}}={}^{50}{{C}_{18-1}}{{2}^{50-(18-1)}}{{a}^{18-1}}={}^{50}{{C}_{17}}{{2}^{33}}{{a}^{17}}\] ………………………………. (6)
But given that in the expansion of\[{{\left( 2+a \right)}^{50}}\], the 17th and ${{17}^{th}}\text{ and }{{18}^{th}}$ term are equal, therefore we get
\[\begin{align}
  & \Rightarrow {}^{50}{{C}_{16}}{{2}^{34}}{{a}^{16}}={}^{50}{{C}_{17}}{{2}^{33}}{{a}^{17}} \\
 & \Rightarrow \dfrac{{{a}^{17}}}{{{a}^{16}}}=\dfrac{{}^{50}{{C}_{16}}{{2}^{34}}}{{}^{50}{{C}_{17}}{{2}^{33}}}
\end{align}\]
\[\Rightarrow a=2\times \dfrac{\dfrac{50!}{(50-16)!16!}}{\dfrac{50!}{(50-17)!17!}}\]
\[\Rightarrow a=2\times \dfrac{\left( 33! \right)\left( 17! \right)}{\left( 34! \right)\left( 16! \right)}\]
Or, \[\Rightarrow a=2\times \dfrac{\left( 33! \right)\times 17\times \left( 16! \right)}{34\times \left( 33! \right)\left( 16! \right)}\]
On simplifying, we get
\[\Rightarrow a=2\times \dfrac{17}{34}\]
On solving, we get
\[\Rightarrow a=1\]

Here we got the value of \[a=1\].

Note: Factorial of a number n that is \[n!\] defined only for any nonnegative integer n. The factorial of number n can be expressed as \[n!=n\left( n-1 \right)\left( n-2 \right)\times ........\times 3\times 2\times 1\] or \[n!=n\left( n-1 \right)!\]. Try not to make any calculation mistakes.