Let P(n) denote the statement that n2 + n is odd. Then,
n > 1
n
n > 2
None of these
P(1):12+1 is odd→not trueP(2):22+2 is odd→not true Suppose P(k) is true. k2+k is odd⇒k2+k=2m+1P(k+1):(k+1)2+(k+1)=(k2+2k+1)+(k+1)=(k2+k)+(2k+2)=2m+1+2k+22(m+k+1)+1⇒P(k+1) is true
Let P(n) denote the statement that n2 + n is odd. It
is seem that P(n) ⇒ P(n + 1), Pn is true for all
Let f(n)=∣∣ ∣ ∣∣nn+1n+2nPnn+1Pn+1n+1Pn+2nCnn+1Cn+1n+1Cn+2∣∣ ∣ ∣∣ Then, f(n) is divisible by