Question
Give an example of a statement P(n) which is true for all n. Justify your answer.
✓
Answer
The required statement is,
$\text{p(n): }1+2+3+\ ....... \ +\text{n}=\frac{\text{n}(\text{n}+1)}{2}$
Justification,
$\text{n}=1,\text{ P(1): }1=\frac{(1+1)}{2}$
Therefore P(1) is true.
Assume $\text{P(k): } 1 + 2 + 3 +\ .....\ +\text{k}=\frac{\text{k}(\text{k}+1)}{2}\ .....(\text{i})$ is true.
Now we have to prove ${\text{P}(\text{k+1): }}1+2+3+\ ....\ +\text{k}+(\text{k}+1)=\frac{(\text{k}+1)(\text{k}+2)}{2}$ is true.
Adding k + 1 on both sides of equation (i) we get,
$1+2+3+\ ....\ +\text{k}+(\text{k}+1)$
$=\frac{\text{k}(\text{k}+1)}{2}+(\text{k}+1)=(\text{k}+1)\Big(\frac{\text{k}}{2}+1\Big)$
$=\frac{(\text{k}+1)(\text{k}+2)}{2}$
$\Rightarrow\text{P}(\text{k}+1)$ is true.
Hence, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have P(n) is true for all n.
$\text{p(n): }1+2+3+\ ....... \ +\text{n}=\frac{\text{n}(\text{n}+1)}{2}$
Justification,
$\text{n}=1,\text{ P(1): }1=\frac{(1+1)}{2}$
Therefore P(1) is true.
Assume $\text{P(k): } 1 + 2 + 3 +\ .....\ +\text{k}=\frac{\text{k}(\text{k}+1)}{2}\ .....(\text{i})$ is true.
Now we have to prove ${\text{P}(\text{k+1): }}1+2+3+\ ....\ +\text{k}+(\text{k}+1)=\frac{(\text{k}+1)(\text{k}+2)}{2}$ is true.
Adding k + 1 on both sides of equation (i) we get,
$1+2+3+\ ....\ +\text{k}+(\text{k}+1)$
$=\frac{\text{k}(\text{k}+1)}{2}+(\text{k}+1)=(\text{k}+1)\Big(\frac{\text{k}}{2}+1\Big)$
$=\frac{(\text{k}+1)(\text{k}+2)}{2}$
$\Rightarrow\text{P}(\text{k}+1)$ is true.
Hence, P(k + 1) is true whenever P(k) is true.
Therefore by the principle of mathematical induction we have P(n) is true for all n.
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