Using the principle of mathematical induction, prove each of the following for all n ϵ N:
$1+2+3+4+\ldots+n=1 / 2 n(n+1)$
To Prove:
$1+2+3+4+\ldots+n=1 / 2 n(n+1)$
Steps to prove by mathematical induction:
Let $P(n)$ be a statement involving the natural number $n$ such that
(i) $P(1)$ is true
(ii) $P(k+1)$ is true, whenever $P(k)$ is true
Then $\mathrm{P}(\mathrm{n})$ is true for all $\mathrm{n} \in \mathrm{N}$
Therefore,
Let $P(n): 1+2+3+4+\ldots+n=1 / 2 n(n+1)$
Step 1:
P(1) = 1/2 1(1 + 1) = 1/2 × 2 = 1
Therefore, P(1) is true
Step 2
Let P(k) is true Then,
P(k): 1 + 2 + 3 + 4 + … + k = 1/2 k(k + 1)
Now,
1 + 2 + 3 + 4 + … + k + (k + 1) = 1/2 k(k + 1) + (k + 1)
= (k + 1){ 1/2 k + 1}
= 1/2 (k + 1) (k + 2)
= P(k + 1)
Hence, $P(k+1)$ is true whenever $P(k)$ is true
Hence, by the principle of mathematical induction, we have
$1+2+3+4+\ldots+n=1 / 2 n(n+1)$ for all $n \in N$
Hence proved.
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