(n^2+ n) is ........... for all n N. - Vidyadip

MCQ

$(n^2+ n)$ is $...........$ for all $n \in N.$

✓
Even
B
Odd
C
Either even or odd
D
None of these

✓

Answer

Correct option: A.

Even

Concept:
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., truth of $P(k)$ implies the truth of $P(k + 1)$
Then, $P(n)$ is true for all natural numbers $n$
Calculation:
Given:
$P(n)=n^2+n$
Put, $\mathrm{n}=1$
$P(1)=12+1=2 ($Even$)$
Let $P(k)$ is true for $n=k$
$P(k):\left(k^2+k\right)$ is even
$\left(k^2+k\right)=2 m$ for some natural number $m$
Now, $P(k+1)=(k+1)^2+(k+1)=k^2+3 k+2=\left(k^2+k\right)+2(k+1)$
using equation $(1), P(k + 1) = 2m + 2(k + 1) = 2[m + (k + 1)],$ which is even
Hence, $P(k +1)$ is even
$\Rightarrow P(k + 1)$ 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 $n \in N$.
i.e $p(n) = (n^2+ n)$ is even

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