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Principle of Mathematical Induction — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsPrinciple of Mathematical Induction1 Mark
MCQ
$P(n): 2 \times 7n + 3 \times 5n - 5$ is divisible by:
  • ✓
    $24, \forall n \in N$
  • B
    $21, \forall n \in N$
  • C
    $32, \forall n \in N$
  • D
    $50, \forall n \in N$
✓

Answer

Correct option: A.
$24, \forall n \in N$
Concepts:
Suppose there is a given statement $P(n)$ involving the natural number $n$ such that
The statement is true for $n = 1,$
i.e., $P(1)$ is true, and
If the statement is true for $n = k ($where $k$ is some positive integer$)$, then the statement is also true for $n = k + 1,$
i.e., the truth of $P(k)$ implies the truth of $P(k + 1).$
Then, $P(n)$ is true for all natural numbers $n$​​​​​​​
Calculation:
Given:
$P(n) = 2 \times 7n + 3 \times 5n - 5$
Put $n = 1$
$P(1) = 2 \times 7^1+ 3 \times 5^1- 5 = 24$, Which is divisible by $24$
Assume $P(k)$ is true
$P(k) = 2 \times 7^k+ 3 \times 5^k- 5 = 24q$, where $q\ \epsilon\ N .....(1)$
Now,
$ T(k+1)=2 \times 7^{k+1}+3 \times 5^{k+1}-5=2 \times 7^k \times 7+3 \times 5^k \times 5-5 $
$ \Rightarrow 7\left\{2 \times 7^k+3 \times 5^k-5-3 \times 5^k+5\right\}+3 \times 5^k \times 5-5 $
$ \Rightarrow 7\left\{24 q-3 \times 5^k+5\right\}+15 \times 5^k-5 $
$ \Rightarrow(7 \times 24 q)-21 \times 5^k+35+15 \times 5^k-5 $
$ \Rightarrow(7 \times 24 q)-6 \times 5^k+30=(7 \times 24 q)-6\left(5^k-5\right) $
$\Rightarrow(7 \times 24 q)-6 \times(4 p)$ As $(5^k-5) $ is a multiple of $4 \}$
$\Rightarrow (7 \times 24q) - 24p = 24(7q - p)$
$\Rightarrow 24 \times r, r = 7q - p,$ is some natural number $.......(2)$
Thus, $P(k + 1)$ is true whenever $P(k)$ is true

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