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Principle of Mathematical Induction — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsPrinciple of Mathematical Induction1 Mark
MCQ
$1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)}:$
  • ✓
    $\{n(n + 3)\}/\{4(n + 1)(n + 2)\}$
  • B
    $(n + 3)/\{4(n + 1)(n + 2)\}$
  • C
    $n/\{4(n + 1)(n + 2)\}$
  • D
    None of these
✓

Answer

Correct option: A.
$\{n(n + 3)\}/\{4(n + 1)(n + 2)\}$
Let $P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/\{n(n + 1)(n + 2)\} = \{n(n + 3)\}/\{4(n + 1)(n + 2)\}$
Putting $n = 1$ in the given statement, we get
$\text{LHS} = 1/(1 ∙ 2 ∙ 3) = 1/6$ and $\text{RHS} = \{1 \times (1 + 3)\}/[4 \times (1 + 1)(1 + 2)] = ( 1 \times 4)/(4 \times 2 \times 3) = 1/6.$
Therefore $\text{LHS = RHS.}$
Thus, the given statement is true for $n = 1,$
i.e., $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + … + 1/\{k(k + 1) (k + 2)\} = \{k(k + 3)\}/\{4(k + 1) (k + 2)\}.…(i)$
Now, $1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/\{k(k + 1) (k + 2)\} + 1/\{(k + 1) (k + 2) (k + 3)\}$
$= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1) (k + 2}] + 1/{(k + 1)(k + 2) (k + 3)}$
$= [\{k(k + 3)\}/\{4(k + 1)(k + 2)\} + 1/\{(k + 1)(k + 2)(k + 3)\}] [$using $(i)]$
$= {k(k + 3)^2 + 4}/\{4(k + 1)(k + 2) (k + 3)\}$
$= (k^3 + 6k^2 + 9k + 4)/\{4(k + 1) (k + 2) (k + 3)\}$
$= {(k + 1) (k + 1) (k + 4)}/\{4 (k + 1) (k + 2) (k + 3)\}$
$= \{(k + 1) (k + 4)\}/\{4(k + 2) (k + 3)\}$
$\Rightarrow P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/\{(k + 1) (k + 2) (k + 3)\}$
$= \{(k + 1) (k + 2)\}/\{4(k + 2) (k + 3)\}$
$\Rightarrow P(k + 1)$ 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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