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Hint:Use the concept of mathematical induction by substituting the values of n as 1, k and k+1. Check whether P(1) holds true or not, if yes then assume then P(k) also holds true, if with this assumption P(k+1) holds true then its proved that ${2.7^n} + {3.5^n} - 5$ is divisible by 24, under given conditions.
Complete step-by-step answer:
Given expression
${2.7^n} + {3.5^n} - 5$
Proof –
Let,
$P\left( n \right) = {2.7^n} + {3.5^n} - 5$ Is divisible by 24.
We note that P (n) is true when n = 1,
Since P (1) = 14 + 15 – 5 = 24 which is divisible by 24.
Assume that it is true for P (k).
I.e. $P\left( k \right) = {2.7^k} + {3.5^k} - 5 = 24q$.................... (1), when $q \in N$
Now according to the mathematical induction principle we have to prove it is also true for P (k + 1) whenever p (k) is true.
Now substitute in place of k, (k + 1) we have,
$ \Rightarrow P\left( {k + 1} \right) = {2.7^{k + 1}} + {3.5^{k + 1}} - 5$
$ \Rightarrow {2.7.7^k} + {3.5.5^k} - 5$
Now add and subtract by ${3.7.5^k} - 7.5$ we have,
$ \Rightarrow {2.7.7^k} + {3.5.5^k} - 5 + {3.7.5^k} - 7.5 - \left( {{{3.7.5}^k} - 7.5} \right)$
$ \Rightarrow 7\left( {{{2.7}^k} + {{3.5}^k} - 5} \right) + {3.5.5^k} - 5 - \left( {{{3.7.5}^k} - 7.5} \right)$
Now from equation (1) we have,
$ \Rightarrow 7\left( {24q} \right) + {3.5.5^k} - 5 - {3.7.5^k} + 7.5$
$ \Rightarrow 7\left( {24q} \right) - {2.3.5^k} - 5 + 35$
$ \Rightarrow 7\left( {24q} \right) - {2.3.5^k} + 30$
$ \Rightarrow 7\left( {24q} \right) - 6\left( {{5^k} - 5} \right)$
Now as we know that $\left( {{5^k} - 5} \right)$ is a multiple of 4 so in place of that we can write (4p) where (p) belongs to the natural number.
$ \Rightarrow 7\left( {24q} \right) - 6\left( {4p} \right)$
\[ \Rightarrow 24\left( {7q - p} \right)\]
So as we see this is a multiple of 24.
Thus P (k + 1) is true whenever P (k) is true.
So according to the principle of mathematical induction ${2.7^n} + {3.5^n} - 5$ is divisible by 24.
Hence proved.
Note – Mathematical induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number. The key point is the condition of natural numbers, so if a proof is to be done and the condition given holds true for natural number only than always think of the mathematical induction
Complete step-by-step answer:
Given expression
${2.7^n} + {3.5^n} - 5$
Proof –
Let,
$P\left( n \right) = {2.7^n} + {3.5^n} - 5$ Is divisible by 24.
We note that P (n) is true when n = 1,
Since P (1) = 14 + 15 – 5 = 24 which is divisible by 24.
Assume that it is true for P (k).
I.e. $P\left( k \right) = {2.7^k} + {3.5^k} - 5 = 24q$.................... (1), when $q \in N$
Now according to the mathematical induction principle we have to prove it is also true for P (k + 1) whenever p (k) is true.
Now substitute in place of k, (k + 1) we have,
$ \Rightarrow P\left( {k + 1} \right) = {2.7^{k + 1}} + {3.5^{k + 1}} - 5$
$ \Rightarrow {2.7.7^k} + {3.5.5^k} - 5$
Now add and subtract by ${3.7.5^k} - 7.5$ we have,
$ \Rightarrow {2.7.7^k} + {3.5.5^k} - 5 + {3.7.5^k} - 7.5 - \left( {{{3.7.5}^k} - 7.5} \right)$
$ \Rightarrow 7\left( {{{2.7}^k} + {{3.5}^k} - 5} \right) + {3.5.5^k} - 5 - \left( {{{3.7.5}^k} - 7.5} \right)$
Now from equation (1) we have,
$ \Rightarrow 7\left( {24q} \right) + {3.5.5^k} - 5 - {3.7.5^k} + 7.5$
$ \Rightarrow 7\left( {24q} \right) - {2.3.5^k} - 5 + 35$
$ \Rightarrow 7\left( {24q} \right) - {2.3.5^k} + 30$
$ \Rightarrow 7\left( {24q} \right) - 6\left( {{5^k} - 5} \right)$
Now as we know that $\left( {{5^k} - 5} \right)$ is a multiple of 4 so in place of that we can write (4p) where (p) belongs to the natural number.
$ \Rightarrow 7\left( {24q} \right) - 6\left( {4p} \right)$
\[ \Rightarrow 24\left( {7q - p} \right)\]
So as we see this is a multiple of 24.
Thus P (k + 1) is true whenever P (k) is true.
So according to the principle of mathematical induction ${2.7^n} + {3.5^n} - 5$ is divisible by 24.
Hence proved.
Note – Mathematical induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number. The key point is the condition of natural numbers, so if a proof is to be done and the condition given holds true for natural number only than always think of the mathematical induction
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