Practice Session - Practice Questions & MCQ

Prove the statement:

$\mathrm{P}(\mathrm{n})=1+2+3+\ldots \ldots+\mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$ is true for all natural numbers.
Proof:
Step 1: Check if $\mathrm{P}(1)$ is true
For $n=1$, we have

$
\begin{array}{ll}
\mathrm{P}(1): & 1=\frac{1(1+1)}{2} \\
\Rightarrow & 1=1
\end{array}
$
So, formula is true for $n=1$. So $P(1)$ is true

Step 2: Assume $P(k)$ to be true and using that check if $P(k+1)$ is true
Suppose that $P(k)$ is true for some positive integer $k$

$
\mathrm{P}(\mathrm{k})=1+2+3+\ldots \ldots+\mathrm{k}=\frac{\mathrm{k}(\mathrm{k}+1)}{2}
$
Now we need to prove that $P(k+1)$ is also true.

$
\mathrm{P}(\mathrm{k}+1)=1+2+3+\ldots \ldots+(\mathrm{k}+1)=\frac{(\mathrm{k}+1)(\mathrm{k}+2)}{2}
$

we already have,

$
1+2+3+\ldots \ldots+k=\frac{k(k+1)}{2}
$

add $(k+1)$ both side in eq (i)

$
\begin{array}{ll}
\Rightarrow & 1+2+3+\ldots \ldots+(k+1)=\frac{k(k+1)}{2}+(k+1) \\
\Rightarrow & 1+2+3+\ldots \ldots+(k+1)=\frac{(k+1)}{2}(k+2)
\end{array}
$
Therefore, $\mathrm{P}(\mathrm{k}+1)$ is true and the inductive proof is now completed.
Hence $P(n)$ is true for all natural numbers $n$.