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Show that ${{3}^{2n+5}}+160{{n}^{2}}-56n-243$ is divisible by $512$.

Last updated date: 14th Aug 2024
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Hint: To prove that the given statement is correct, use induction on $n$. Check the validity of the statement for $n=1$ and then assuming that the statement holds for $n=k$, prove the statement for $n=k+1$.

We have the statement that ${{3}^{2n+5}}+160{{n}^{2}}-56n-243$ is divisible by $512$. We have to prove this statement. We will do so by using induction on $n$. We will check the validity of the given statement for $n=1$ and then assuming that the statement holds for $n=k$, we will prove the statement for $n=k+1$ .
Thus, substituting $n=1$ in the equation ${{3}^{2n+5}}+160{{n}^{2}}-56n-243$, we get${{3}^{2+5}}+160\left( 1 \right)-56\left( 1 \right)-243={{3}^{7}}+160-56-243=2187-139=2048$.
We observe that $\dfrac{2048}{512}=4$.
Hence, we observe that the statement ${{3}^{2n+5}}+160{{n}^{2}}-56n-243$ is divisible by $512$ holds for $n=1$.
We will now prove the statement for $n=k+1$, assuming that the statement holds for $n=k$.
As the given statement holds for $n=k$, we will replace $n$ by $k$ in the statement that ${{3}^{2n+5}}+160{{n}^{2}}-56n-243$ is divisible by $512$.
Replacing $n$ by $k$, we get ${{3}^{2k+5}}+160{{k}^{2}}-56k-243$ is divisible by $512$.
As ${{3}^{2k+5}}+160{{k}^{2}}-56k-243$ is divisible by $512$, we can write it as ${{3}^{2k+5}}+160{{k}^{2}}-56k-243=512a$, where $a$ represents some integer.
We will now prove that the statement holds for $n=k+1$.
Substituting $n=k+1$in the expression ${{3}^{2n+5}}+160{{n}^{2}}-56n-243$, we get ${{3}^{2\left( k+1 \right)+5}}+160{{\left( k+1 \right)}^{2}}-56\left( k+1 \right)-243$.
Simplifying the above expression, we have ${{3}^{2k+5}}\times 9+160\left( {{k}^{2}}+2k+1 \right)-56k-56-243$.
\begin{align} & \Rightarrow {{3}^{2k+5}}\left( 1+8 \right)+160{{k}^{2}}+160+320k-56k-56-243 \\ & \Rightarrow \left( {{3}^{2k+5}}+160{{k}^{2}}-56k-243 \right)+\left( 8\times {{3}^{2k+5}}+160+320k-56 \right) \\ & \Rightarrow 512a+8\times {{3}^{2k+5}}+104+320k \\ & \Rightarrow 512a+8\left( {{3}^{2k+5}}+40k+13 \right) \\ \end{align}
Adding and subtracting the term $8\left( 160{{k}^{2}}-56k-243 \right)$ from the above expression, we have $512a+8\left( {{3}^{2k+5}}+160{{k}^{2}}-56k-243-160{{k}^{2}}+56k+243+40k+13 \right)$.
\begin{align} & \Rightarrow 512a+8\left( {{3}^{2k+5}}+160{{k}^{2}}-56k-243 \right)+8\left( -160{{k}^{2}}+96k+256 \right) \\ & \Rightarrow 512a+8\left( 512a \right)-8\times 32\left( 5{{k}^{2}}-3k-8 \right) \\ & \Rightarrow 512b-256\left( 5{{k}^{2}}-3k-8 \right) \\ \end{align}
If we substitute $k=1$ in the above expression, we get $512b-256\left( -6 \right)=512\left( b+3 \right)=512c$ where $b$ and $c$ are some integers.
Thus, we observe that we can write $512b-256\left( 5{{k}^{2}}-3k-8 \right)=512c$ for integer $c$.
Hence, we have ${{3}^{2\left( k+1 \right)+5}}+160{{\left( k+1 \right)}^{2}}-56\left( k+1 \right)-243=512c$ thus, proving that the given statement is true.
Thus, the statement ${{3}^{2n+5}}+160{{n}^{2}}-56n-243$ is divisible by $512$ holds for all $n$.

Note: We can also prove this statement by using induction in another way by assuming that the given statement holds for $n-1$ and then proving it for $n$.