Question

Using binomial theorem prove that ${{3}^{2n+2}}-8n-9$ is divisible by 64, $n \in N
$

Answer 1

Hint: Binomial theorem states that for any positive integer n, the nth power of the sum of two numbers a and b may be expressed as the sum of n+1 terms of the form. This theorem is generally useful in algebra as well as for determining permutations and combinations and probabilities.

Complete step-by-step solution:
The binomial theorem is
\[{{\left( x+y \right)}^{n}}=\sum\limits_{k=0}^{n}{\left( {}^{n}{{C}_{k}} \right){{x}^{n-k}}{{y}^{k}}=}\sum\limits_{k=0}^{n}{\left( {}^{n}{{C}_{k}} \right){{x}^{k}}{{y}^{n-k}}}\]
Binomial coefficient of each term =\[\left( {}^{n}{{C}_{k}} \right)=\dfrac{n!}{\left( n-k \right)!k!}\]
Here, we have to prove that ${{3}^{2n+2}}-8n-9$ is divisible by 64 which means ${{3}^{2n+2}}-8n-9$ should be a multiple of 64.
Let us assume that A=${{3}^{2n+2}}-8n-9$
$\begin{align}
  &\Rightarrow A={{\left( 3 \right)}^{2\left( n+1 \right)}}-8n-9 \\
 &\Rightarrow A={{\left( {{3}^{2}} \right)}^{n+1}}-8n-9 \\
 &\Rightarrow A={{\left( 9 \right)}^{n+1}}-8n-9 \\
 &\Rightarrow A+8n+9={{\left( 9 \right)}^{n+1}} \\
\end{align}$
Here, ${{\left( 9 \right)}^{n+1}}$ can be expanded using binomial theorem
${{\left( 9 \right)}^{n+1}}={{\left( 1+8 \right)}^{n+1}}$ Comparing it with \[{{\left( x+y \right)}^{n}}=\sum\limits_{k=0}^{n}{\left( {}^{n}{{C}_{k}} \right){{x}^{n-k}}{{y}^{k}}}\]
We have x = 1, y = 8, substituting the values in the above equation we get
\[\begin{align}
& {{\left( 1+8 \right)}^{n+1}}=\sum\limits_{k=0}^{n+1}{\left( {}^{n+1}{{C}_{k}} \right){{1}^{n-k}}{{8}^{k}}} \\
 & \text{ =}\sum\limits_{k=0}^{n+1}{\left( {}^{n+1}{{C}_{k}} \right){{8}^{k}}} \\
 & \text{Expanding the summation we get} \\
&\Rightarrow {{\left( 1+8 \right)}^{n+1}}=\left( {}^{n+1}{{C}_{0}} \right){{8}^{0}}+\left( {}^{n+1}{{C}_{1}} \right){{8}^{1}}+\left( {}^{n+1}{{C}_{2}} \right){{8}^{2}}+\left( {}^{n+1}{{C}_{3}} \right){{8}^{3}}+............\left( {}^{n+1}{{C}_{n}} \right){{8}^{n}}+\left( {}^{n+1}{{C}_{n+1}} \right){{8}^{n+1}} \\
\end{align}\]
Here , ${}^{n+1}{{C}_{0}}=\dfrac{(n+1)!}{(n+1-0)!0!}=1,{}^{n+1}{{C}_{1}}=\dfrac{(n+1)!}{(n+1-1)!1!}=n+1\text{ and so on}$
Substituting the values of the above binomial coefficients in the expansion we get
\[\begin{align}
  & {{\left( 1+8 \right)}^{n+1}}=\left( 1 \right){{8}^{0}}+\left( n+1 \right){{8}^{1}}+\left( {}^{n+1}{{C}_{2}} \right){{8}^{2}}+\left( {}^{n+1}{{C}_{3}} \right){{8}^{2}}{{.8}^{1}}+............\left( {}^{n+1}{{C}_{n}} \right){{8}^{n-2}}{{.8}^{2}}+\left( {}^{n+1}{{C}_{n+1}} \right){{8}^{n-1}}{{.8}^{2}} \\
 &\Rightarrow {{9}^{n+1}}=1+8(n+1)+{{8}^{2}}\left[ {}^{n+1}{{C}_{2}}+...........\left( {}^{n+1}{{C}_{n}} \right){{8}^{n-2}}+\left( {}^{n+1}{{C}_{n+1}} \right){{8}^{n-1}} \right] \\
 &\Rightarrow {{9}^{n+1}}=1+8n+8+64(\text{positive number as binomial coeffecient wouldn }\!\!'\!\!\text{ t be negative)} \\
 &\Rightarrow {{9}^{n+1}}=8n+9+64(\text{positive number)} \\
\end{align}\]
As assumed before $A+8n+9={{\left( 9 \right)}^{n+1}}$
$A + 8n + 9 = 8n + 9 +64$(positive number)
$\Rightarrow A = 64$(positive number)
$\Rightarrow {{3}^{2n+2}}-8n-9= 64$(positive number)
Therefore, it can be interpreted from the above statement that ${{3}^{2n+2}}-8n-9$ is divisible by 64.

Note: If observed carefully we have here expanded 9 as 1 + 8 and made our x=1 and y=9 but if we had expanded it as 9=8+1 then we will have x=8 and y=1 and substitute it in the formula \[\sum\limits_{k=0}^{n}{\left( {}^{n}{{C}_{k}} \right){{x}^{n-k}}{{y}^{k}}}\] as x=8 while summing up the values we start with higher powers of 8 like ${{8}^{n}},{{8}^{n-1}},{{8}^{n-2}},........,{{8}^{n-(n-1)}},{{8}^{n-n}}$ but for our question we need lower powers of 8 like at the first just to not end up in confusion. So it would be easy to solve if we choose our variables accordingly.