
Show that ${{8.7}^{n}}+{{4}^{n+2}}$ is of the form $24\left( 2r-1 \right)$.
Answer
560.7k+ views
Hint: To solve this question we will use the concept of mathematical induction. Let us assume that $P(n)={{8.7}^{n}}+{{4}^{n+2}}$ is true. Then by putting the value $n=1$ and $n=n+1$ we will check that the equation is true or not then solve the equation and convert it in the form $24\left( 2r-1 \right)$.
Complete step by step answer:
We have been given that ${{8.7}^{n}}+{{4}^{n+2}}$.
We have to show that ${{8.7}^{n}}+{{4}^{n+2}}$ is of the form $24\left( 2r-1 \right)$.
We have given ${{8.7}^{n}}+{{4}^{n+2}}$
Let $P(n)={{8.7}^{n}}+{{4}^{n+2}}$
Let us put the value $n=1$ we get
$\Rightarrow P(1)={{8.7}^{1}}+{{4}^{1+2}}$
Now simplifying further we get
$\begin{align}
& \Rightarrow P(1)=8.7+{{4}^{3}} \\
& \Rightarrow P(1)=56+64 \\
& \Rightarrow P(1)=120 \\
\end{align}$
Therefore we can write as
$\begin{align}
& \Rightarrow P(1)=24\times 5 \\
& \Rightarrow P(1)=24\left( 2\left( 3 \right)-1 \right) \\
\end{align}$
Which is of the form $24\left( 2r-1 \right)$.
Now, Let us put the value $n=n+1$ we will get
$P(n+1)={{8.7}^{n+1}}+{{4}^{n+1+2}}$
Now, simplifying further we will get
$\Rightarrow P(n+1)={{8.7}^{n+1}}+{{4}^{n+3}}$
Now, let us rewrite the above obtained equation as
$\Rightarrow P(n+1)-P(n)={{8.7}^{n+1}}+{{4}^{n+3}}-\left( {{8.7}^{n}}+{{4}^{n+2}} \right)$
Now, simplifying further the above obtained equation we will get
$\Rightarrow P(n+1)-P(n)={{8.7}^{n+1}}-{{8.7}^{n}}+{{4}^{n+3}}-{{4}^{n+2}}$
Now, taking the common terms out we will get
$\Rightarrow P(n+1)-P(n)={{8.7}^{n}}\left( 7-1 \right)+{{4}^{n+2}}\left( 4-1 \right)$
Now, simplifying further the above obtained equation we will get
$\begin{align}
& \Rightarrow P(n+1)-P(n)={{8.7}^{n}}\times 6+{{4}^{n+2}}\times 3 \\
& \Rightarrow P(n+1)-P(n)={{48.7}^{n}}+{{4}^{n}}\times {{4}^{2}}\times 3 \\
& \Rightarrow P(n+1)-P(n)={{48.7}^{n}}+{{48.4}^{n}} \\
& \Rightarrow P(n+1)-P(n)=48\left( {{7}^{n}}+{{4}^{n}} \right) \\
\end{align}$
Now, in the above obtained equation ${{7}^{n}}$ is always odd and ${{4}^{n}}$ is always even but the sum of both terms is always odd so it is of the form $2r-1$.
So we can say that $P(n+1)$ is also of the form $24\left( 2r-1 \right)$.
So we can say that ${{8.7}^{n}}+{{4}^{n+2}}$ is of the form $24\left( 2r-1 \right)$.
Hence proved
Note: Mathematical induction is a technique of proving a statement true for all natural numbers. Mathematical induction consists of three steps to prove a statement true or false. Alternatively one way of looking at mathematical induction is to prove not the one proposition but a whole sequence of prepositions.
Complete step by step answer:
We have been given that ${{8.7}^{n}}+{{4}^{n+2}}$.
We have to show that ${{8.7}^{n}}+{{4}^{n+2}}$ is of the form $24\left( 2r-1 \right)$.
We have given ${{8.7}^{n}}+{{4}^{n+2}}$
Let $P(n)={{8.7}^{n}}+{{4}^{n+2}}$
Let us put the value $n=1$ we get
$\Rightarrow P(1)={{8.7}^{1}}+{{4}^{1+2}}$
Now simplifying further we get
$\begin{align}
& \Rightarrow P(1)=8.7+{{4}^{3}} \\
& \Rightarrow P(1)=56+64 \\
& \Rightarrow P(1)=120 \\
\end{align}$
Therefore we can write as
$\begin{align}
& \Rightarrow P(1)=24\times 5 \\
& \Rightarrow P(1)=24\left( 2\left( 3 \right)-1 \right) \\
\end{align}$
Which is of the form $24\left( 2r-1 \right)$.
Now, Let us put the value $n=n+1$ we will get
$P(n+1)={{8.7}^{n+1}}+{{4}^{n+1+2}}$
Now, simplifying further we will get
$\Rightarrow P(n+1)={{8.7}^{n+1}}+{{4}^{n+3}}$
Now, let us rewrite the above obtained equation as
$\Rightarrow P(n+1)-P(n)={{8.7}^{n+1}}+{{4}^{n+3}}-\left( {{8.7}^{n}}+{{4}^{n+2}} \right)$
Now, simplifying further the above obtained equation we will get
$\Rightarrow P(n+1)-P(n)={{8.7}^{n+1}}-{{8.7}^{n}}+{{4}^{n+3}}-{{4}^{n+2}}$
Now, taking the common terms out we will get
$\Rightarrow P(n+1)-P(n)={{8.7}^{n}}\left( 7-1 \right)+{{4}^{n+2}}\left( 4-1 \right)$
Now, simplifying further the above obtained equation we will get
$\begin{align}
& \Rightarrow P(n+1)-P(n)={{8.7}^{n}}\times 6+{{4}^{n+2}}\times 3 \\
& \Rightarrow P(n+1)-P(n)={{48.7}^{n}}+{{4}^{n}}\times {{4}^{2}}\times 3 \\
& \Rightarrow P(n+1)-P(n)={{48.7}^{n}}+{{48.4}^{n}} \\
& \Rightarrow P(n+1)-P(n)=48\left( {{7}^{n}}+{{4}^{n}} \right) \\
\end{align}$
Now, in the above obtained equation ${{7}^{n}}$ is always odd and ${{4}^{n}}$ is always even but the sum of both terms is always odd so it is of the form $2r-1$.
So we can say that $P(n+1)$ is also of the form $24\left( 2r-1 \right)$.
So we can say that ${{8.7}^{n}}+{{4}^{n+2}}$ is of the form $24\left( 2r-1 \right)$.
Hence proved
Note: Mathematical induction is a technique of proving a statement true for all natural numbers. Mathematical induction consists of three steps to prove a statement true or false. Alternatively one way of looking at mathematical induction is to prove not the one proposition but a whole sequence of prepositions.
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