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Question: Using binomial theorem prove that \({{3}^{2n+2}}-8n-9\) is divisible by 64, \(n \in N \)...

Using binomial theorem prove that 32n+2−8n−9{{3}^{2n+2}}-8n-9 is divisible by 64, n∈Nn \in N

Explanation

Solution

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
(x+y)n=∑k=0n(nCk)xn−kyk=∑k=0n(nCk)xkyn−k{{\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 =(nCk)=n!(n−k)!k!\left( {}^{n}{{C}_{k}} \right)=\dfrac{n!}{\left( n-k \right)!k!}
Here, we have to prove that 32n+2−8n−9{{3}^{2n+2}}-8n-9 is divisible by 64 which means 32n+2−8n−9{{3}^{2n+2}}-8n-9 should be a multiple of 64.
Let us assume that A=32n+2−8n−9{{3}^{2n+2}}-8n-9
⇒A=(3)2(n+1)−8n−9 ⇒A=(32)n+1−8n−9 ⇒A=(9)n+1−8n−9 ⇒A+8n+9=(9)n+1 \begin{aligned} &\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{aligned}
Here, (9)n+1{{\left( 9 \right)}^{n+1}} can be expanded using binomial theorem
(9)n+1=(1+8)n+1{{\left( 9 \right)}^{n+1}}={{\left( 1+8 \right)}^{n+1}} Comparing it with (x+y)n=∑k=0n(nCk)xn−kyk{{\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

& {{\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{aligned}$$ Here , ${}^{n+1}{{C}_{0}}=\dfrac{(n+1)!}{(n+1-0)!0!}=1,{}^{n+1}{{C}_{1}}=\dfrac{(n+1)!}{(n+1-1)!1!}=n+1\text{ and so on}$ Substituting the values of the above binomial coefficients in the expansion we get $$\begin{aligned} & {{\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{aligned}$$ As assumed before $A+8n+9={{\left( 9 \right)}^{n+1}}$ $A + 8n + 9 = 8n + 9 +64$(positive number) $\Rightarrow A = 64$(positive number) $\Rightarrow {{3}^{2n+2}}-8n-9= 64$(positive number) **Therefore, it can be interpreted from the above statement that ${{3}^{2n+2}}-8n-9$ is divisible by 64.** **Note:** If observed carefully we have here expanded 9 as 1 + 8 and made our x=1 and y=9 but if we had expanded it as 9=8+1 then we will have x=8 and y=1 and substitute it in the formula $$\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.