1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1):

MCQ

$1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1):$

A
$n(n + 1)(n + 2)$
B
$\{n(n + 1)(n + 2)\}/2$
✓
$\{n(n + 1)(n + 2)\}/3$
D
$\{n(n + 1)(n + 2)\}/4$

✓

Answer

Correct option: C.

$\{n(n + 1)(n + 2)\}/3$

Let the given statement be $P(n).$ Then,
$P(n): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + n(n + 1) = (1/3)\{n(n + 1) (n + 2)\}$
Thus, the given statement is true for $n = 1,$
i.e., $P(1)$ is true.
Let $P(k)$ be true. Then,
$P(k): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ….. + k(k + 1) = (1/3)\{k(k + 1) (k + 2)\}.$
Now, $1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +…+ k(k + 1) + (k + 1) (k + 2)$
$= (1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + ……. + k(k + 1)) + (k + 1) (k + 2)$
$= (1/3) k(k + 1) (k + 2) + (k + 1)(k + 2) [$using $(i)]$
$= (1/3) [k(k + 1) (k + 2) + 3(k + 1)(k + 2)$
$= (1/3)\{(k + 1) (k + 2)(k + 3)\}$
$\Rightarrow P(k + 1): 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 +……+ (k + 1) (k + 2)$
$= (1/3)\{k + 1 )(k + 2) (k +3)\}$
$\Rightarrow P(k + 1)$ 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 values of $\in N$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Principle of Mathematical Induction→Principle of Mathematical Induction — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Principle of Mathematical Induction — Maths STD 11 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions