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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 \[n\].

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 \[n\].

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