P ( n ) : 1.2 + 2.3 + 3.4 + . . . + n ( n + 1 ) = ( n + 1 ) ( n + 2 ) 3 The statement P(n) is Options: A . &nbsp is true for all natural numbers B . &nbsp is not true for all natural numbers C . &nbsp is true for n=1 D . &nbsp is true for n=2

Answer: Option B : B P ( n ) : 1.2 + 2.3 + 3.4 + . . . + n ( n + 1 ) = ( n + 1 ) ( n + 2 ) 3 P ( 1 ) : 1.2 = 2.3 3 ⇒ 2 = 2 − t r u e Assume P(k) is true ⇒ 1.2 + 2.3 + 3.4 + . . . + k ( k + 1 ) = ( k + 1 ) ( k + 2 ) 3 N o w , P ( k + 1 ) : 1.2 + 2.3 + 3.4 + . . . + k ( k + 1 ) + ( k + 1 ) ( k + 2 ) = ( k + 1 ) ( k + 2 ) 3 + ( k + 1 ) ( k + 2 ) = 4 ( k + 1 ) ( k + 2 ) 3 ≠ ( k + 2 ) ( k + 3 ) 3 P(k+1) is not true. Hence, P(n) is not true for all natural numbers

P(n):1.2+2.3+3.4+...+n(n+1)=(n+1)(n+2)3 The statement P(n) is

Question

$P (n) : 1.2 + 2.3 + 3.4 + . . . + n (n + 1) = \frac{(n + 1) (n + 2)}{3}$
The statement P(n) is

Options:

A . is true for all natural numbers

B . is not true for all natural numbers

C . is true for n=1

D . is true for n=2

Answer: Option B
:
B

$P (n) : 1.2 + 2.3 + 3.4 + . . . + n (n + 1) = \frac{(n + 1) (n + 2)}{3} P (1) : 1.2 = \frac{2.3}{3} \Rightarrow 2 = 2 - t r u e$
Assume P(k) is true
$\Rightarrow 1.2 + 2.3 + 3.4 + . . . + k (k + 1) = \frac{(k + 1) (k + 2)}{3} N o w, P (k + 1) : 1.2 + 2.3 + 3.4 + . . . + k (k + 1) + (k + 1) (k + 2) = \frac{(k + 1) (k + 2)}{3} + (k + 1) (k + 2) = \frac{4 (k + 1) (k + 2)}{3} \neq \frac{(k + 2) (k + 3)}{3}$
P(k+1) is not true.
Hence, P(n) is not true for all natural numbers