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MAVERICK [17]
3 years ago
9

Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks​

Mathematics
1 answer:
Iteru [2.4K]3 years ago
4 0

Let

P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

P(1):\ 1 = \dfrac{1\cdot 2}{2}=1

So, the base case is ok. Now, we need to assume P(n) and prove P(n+1).

P(n+1) states that

P(n+1):\ 1+2+\ldots+n+(n+1) = \dfrac{(n+1)(n+2)}{2}=\dfrac{n^2+3n+2}{2}

Since we're assuming P(n), we can substitute the sum of the first n terms with their expression:

\underbrace{1+2+\ldots+n}_{P(n)}+n+1 = \dfrac{n(n+1)}{2}+n+1=\dfrac{n(n+1)+2n+2}{2}=\dfrac{n^2+3n+2}{2}

Which terminates the proof, since we showed that

P(n+1):\ 1+2+\ldots+n+(n+1) =\dfrac{n^2+3n+2}{2}

as required

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2 years ago
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