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Principle of Mathematical Induction — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsPrinciple of Mathematical Induction1 Mark
MCQ
$(2 ∙ 7^N+ 3 ∙ 5^N– 5)$ is divisible by $………..$ for all $N \in N:$
  • A
    $6$
  • B
    $12$
  • C
    $18$
  • ✓
    $24$
✓

Answer

Correct option: D.
$24$
Let $P(n): (2 ∙ 7^n+ 3 ∙ 5^n– 5)$ is divisible by $24.$
For $n = 1,$ the given expression becomes $(2 ∙ 7^1+ 3 ∙ 5^1– 5) = 24$,
which is clearly divisible by $24.$
So, the given statement is true for $n = 1$,
i.e., $P(1)$ is true.
Let $P(k)$ be true. Then,
$ P(k):\left(2 \cdot 7^n+3 \cdot 5^n-5\right)$ is divisible by $24 $
$ \Rightarrow\left(2 \cdot 7^n+3 \cdot 5^n-5\right)=24 m$, for $m=N $
$ \text { Now, }\left(2 \cdot 7^n+3 \cdot 5^n-5\right) $
$ =\left(2 \cdot 7^k \cdot 7+3 \cdot 5^k \cdot 5-5\right) $
$ =7\left(2 \cdot 7^k+3 \cdot 5^k-5\right)$
$=6 \cdot 5^k+30 $
$ =(7 \times 24 m)-6\left(5^k-5\right) $
$ =(24 \times 7 m)-6 \times 4 p,$
where $\left(5^k-5\right)=5\left(5^{k-1}-1\right)=4 p$
$[$Since $\left(5^{k-1}-1\right)$ is divisible by $(5-1) ]$
$= 24 \times (7m – p)$
$= 24r,$ where $r = (7m – p) \in N$
$\Rightarrow P (k + 1): (2 ∙ 7^k+ 13 ∙ 5^k+ 1 – 5)$ is divisible by $24.$
$\Rightarrow P(k + 1)$ is true, whenever $P(k)$ is true.
Thus, $P(1)$ is true and $P(k + 1)$ is true, whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, $P(n)$ is true for all $n \in N.$

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