Question

Prove that ${{5}^{2\left( n+1 \right)}}-24n-25$ is divisible by 576.

Answer 1

Hint: We are asked to prove that ${{5}^{2\left( n+1 \right)}}-24n-25$ is divisible by 576. For all n, we use induction in which we firstly show that this statement is true for n = 1 then we consider it holds true for n = k. Then we claim that, for $n=k+1$, the statement defined by us is true. To simplify and solve we use Euclid division lemma method, that says, if $\dfrac{a}{b}$ then $b=ay$ and we use ${{x}^{a}}\cdot {{x}^{b}}={{x}^{a+b}}$

Complete step by step answer:
We are given an equation as \[{{5}^{2\left( n+1 \right)}}-24n-25\]
We have to show that, for all $n\in \text{integer}$ the number obtained from this equation is divisible by 576.
We will use mathematical induction. In this, we check that, is the given equation follow of n =1 then we consider that the given equation holds true first k terms and lastly we claim that, it is also true for $n=k+1$ term.
Now, as $k+1$ was arbitrary so, it will tell us that this is true for n. Hence, for all n.
Now, we have ${{5}^{2\left( n+1 \right)}}-24n-25$ we substitute n = 1, we get:
\[\Rightarrow {{5}^{2\left( 1+1 \right)}}-24\left( 1 \right)-25\]
Simplify, we get:
\[\Rightarrow {{5}^{4}}-24-25\]
We know, ${{5}^{4}}=625$ so we get:
\[\Rightarrow 625-24-25=576\]
Simplifying we get, for n = 1 we have:
\[\Rightarrow {{5}^{2\left( n+1 \right)}}-24n-25=576\]
So clearly it is divisible by 576.
Hence, for n = 1 our statement that ${{5}^{2\left( n+1 \right)}}-24n-25$ is divisible by 576 holds true.
Now, let us assume that, the statement is true for some $n=k$
So putting $n=k$ we get:
${{5}^{2\left( k+1 \right)}}-24k-25$ is divisible by 576
So it means, \[{{5}^{2\left( k+1 \right)}}-24k-25=576\times y\cdots \cdots \cdots \left( 1 \right)\]
Now, we will try to claim that, our statement holds true for $n=k+1$
So, we put $n=k+1$ we will get:
\[\Rightarrow {{5}^{2\left( k+1+1 \right)}}-24\left( k+1 \right)-25\]
Now, simplifying we get:
\[\Rightarrow {{5}^{2\left( k+2 \right)}}-24k-25-24\cdots \cdots \cdots \left( 2 \right)\]
Now using equation (1), we get:
\[\Rightarrow -24k-25=576y-{{5}^{2\left( k+1 \right)}}\]
So, using this in equation (2) we get:
\[\begin{align}
  & {{5}^{2\left( k+2 \right)}}+576y-{{5}^{2\left( k+1 \right)}}-24 \\
 & \Rightarrow {{5}^{2\left( k+2 \right)}}=576y+24k+25 \\
\end{align}\]
Using this in equation (2) we get:
\[\Rightarrow {{5}^{2\left( k+1 \right)}}\cdot {{5}^{2}}-24k-25-24=\left( 576y+24k+25 \right){{5}^{2}}-24k-49\]
Opening bracket we get:
\[\begin{align}
  & \Rightarrow 576\cdot {{5}^{2}}y+600k+25\times {{5}^{2}}-24k-49 \\
 & \Rightarrow {{5}^{2}}\cdot 576y+576k+576 \\
\end{align}\]
Taking 576 common, we get:
\[\Rightarrow 576\left( {{5}^{2}}y+k+1 \right)\]
Let $\left( {{5}^{2}}y+k+1 \right)$ as t, so we get:
\[\Rightarrow 576t\]
Now solving further, we get:
\[\Rightarrow {{5}^{2\left( k+1+1 \right)}}-24\left( k+1 \right)-25=576t\]
Where t = $\left( {{5}^{2}}y+k+1 \right)$
So clearly it is divisible by 576.
Hence, our statement is true even for $n=k+1$
As n=k was arbitrary. Hence, this statement is true for any n. So we get:
${{5}^{2\left( n+1 \right)}}-24n-25$ is divisible by 576.

Note: While switching the terms using the digit from n=k step, we need to be very precise as mostly error happens their only. Also remember, ${{x}^{a}}\cdot {{x}^{b}}={{x}^{a+b}}$ so using to this we got $\Rightarrow {{5}^{2\left( k+1+1 \right)}}={{5}^{2\left( k+1 \right)}}\cdot {{5}^{2}}$ as \[\Rightarrow 2\left( k+1+1 \right)=2k+2+2=2\left( k+1 \right)+2\]
Whenever one thing is divisible by other than by Euclid lemma it will be written as multiple.
Say if $\dfrac{a}{b}$ then $b=ay$ for some y.
Remember, if the statement is not valid for n = 1 then we will terminate our procedure there and we get the statement is false.