MCQ
For all $n \in N, 3.5^{2n+1}+ 2^{3n+1}$ is divisible by:
- A$19$
- ✓$17$
- C$23$
- D$25$
✓
Answer
Correct option: B.
$17$
Let $P(n)$ be the statement that $3.5^{2n+1} + 2^{3n+1} $ is divisible by $17$
If $n = 1,$ then given expression $= 3 \times 5^3+ 2^4+ 375 + 16 = 391 = 17 \times 23$, divisible by $17.$
$P(1)$ is true
Assume that $P(k)$ is true.
$3.5^{2k+1}+2^{3k+1}$ is divisible by $17 .$
$ 3.5^{2k}=1+2^{3k+1}=17 {~m}$ where $m \in N$
$ 3.5^{2(k+1)+1}+23^{(k+1)+1}$
$ =3.5^{2k+1} \times 5^2+2^{3k+1} \times 2^3$
$ =25^{(17m-23k+1)}+8.2^{3k+1}$
$ =425m-25.2^{3k+1}+8.2^{3k+1}$
$=425m-17.2^{3k+1} $
$ =17\left(25 \mathrm{~m}-2^{3 \mathrm{k}+1}\right)$, divisible by $17$
$P(k + 1)$ is true by Principle of Mathematical Induction
$P(n)$ is true for all $n \in N. 3.5^{2n+1}+ 2^{3n+1}$ is divisible by $17$ for all $n \in N$
If $n = 1,$ then given expression $= 3 \times 5^3+ 2^4+ 375 + 16 = 391 = 17 \times 23$, divisible by $17.$
$P(1)$ is true
Assume that $P(k)$ is true.
$3.5^{2k+1}+2^{3k+1}$ is divisible by $17 .$
$ 3.5^{2k}=1+2^{3k+1}=17 {~m}$ where $m \in N$
$ 3.5^{2(k+1)+1}+23^{(k+1)+1}$
$ =3.5^{2k+1} \times 5^2+2^{3k+1} \times 2^3$
$ =25^{(17m-23k+1)}+8.2^{3k+1}$
$ =425m-25.2^{3k+1}+8.2^{3k+1}$
$=425m-17.2^{3k+1} $
$ =17\left(25 \mathrm{~m}-2^{3 \mathrm{k}+1}\right)$, divisible by $17$
$P(k + 1)$ is true by Principle of Mathematical Induction
$P(n)$ is true for all $n \in N. 3.5^{2n+1}+ 2^{3n+1}$ is divisible by $17$ for all $n \in N$
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