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Present version from Mathematical induction article [ edit ] Algebra will now establish that
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{\displaystyle {\frac {n(n+1)}{2}}+(n+1)={\frac {(n+1)((n+1)+1)}{2}}\,,}
(Click "show" at right to see the algebraic details or "hide" to hide them.)
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{\displaystyle {\begin{aligned}{\frac {n(n+1)}{2}}+(n+1)&=(n+1)\left({\frac {n}{2}}+1\right)\\&=(n+1)\left({\frac {n}{2}}+{\frac {2}{2}}\right)\\&=(n+1)\left({\frac {n+2}{2}}\right)\\&={\frac {(n+1)(n+2)}{2}}\\&={\frac {(n+1)((n+1)+1)}{2}}.\end{aligned}}}
thereby showing that indeed P (n + 1)
Script version [ edit ] Algebra will now establish that
n
(
n
+
1
)
2
+
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n
+
1
)
=
(
n
+
1
)
(
(
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+
1
)
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,
{\displaystyle {\frac {n(n+1)}{2}}+(n+1)={\frac {(n+1)((n+1)+1)}{2}}\,,}
((n+1)n)/2 + (n+1) = ((n+1)((n+1)+1))/2
((n+1)n)/2 + (n+1) = (n+1)(n/2 + 1)
(n+1)(n/2 + 1) = (n+1)((n+2)/2)
(n+1)((n+2)/2) = ((n+1)(n+2))/2)
((n+1)(n+2))/2) = ((n+1)((n+1) + 1))/2)
thereby showing that indeed P (n + 1)
Algebra will now establish that
n
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n
+
1
)
2
+
(
n
+
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=
(
n
+
1
)
(
(
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+
1
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{\displaystyle {\frac {n(n+1)}{2}}+(n+1)={\frac {(n+1)((n+1)+1)}{2}}\,,}
[derivation 1] thereby showing that indeed P (n + 1)[1]
Algebra will now establish that
n
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+
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=
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1
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+
1
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2
,
{\displaystyle {\frac {n(n+1)}{2}}+(n+1)={\frac {(n+1)((n+1)+1)}{2}}\,,}
[derivation 2] thereby showing that indeed P (n + 1)
Interpolated material [ edit ]
x 2 4
Material with references [ edit ] In 370 BC, Plato ’s dialog Parmenides may have contained the first inductive proof ever.[2] Euclid 's proof that the number of primes is infinite and in Bhaskara 's "cyclic method ".[3] [4]
References [ edit ]
^ This is a normal inline citation.
^ Mathematical Induction: The Basis Step of Verification and Validation in a Modeling and Simulation Course ^ Cajori (1918), p. 197"The process of reasoning called "Mathematical Induction" has had several independent origins. It has been traced back to the Swiss Jakob (James) Bernoulli, the Frenchman B. Pascal and P. Fermat, and the Italian F. Maurolycus. [...] By reading a little between the lines one can find traces of mathematical induction still earlier, in the writings of the Hindus and the Greeks, as, for instance, in the "cyclic method" of Bhaskara, and in Euclid's proof that the number of primes is infinite."
^ Historical reference.
Derivations [ edit ]
^
Derivation of induction formula for summing consecutive positive integers:
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{\displaystyle {\begin{aligned}{\frac {n(n+1)}{2}}+(n+1)&=(n+1)\left({\frac {n}{2}}+1\right)\\&=(n+1)\left({\frac {n}{2}}+{\frac {2}{2}}\right)\\&=(n+1)\left({\frac {n+2}{2}}\right)\\&={\frac {(n+1)(n+2)}{2}}\\&={\frac {(n+1)((n+1)+1)}{2}}.\end{aligned}}}
^
Derivation of induction formula for summing consecutive positive integers:
(Click "show" at right to see the derivation or "hide" to hide it.)
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{\displaystyle {\begin{aligned}{\frac {n(n+1)}{2}}+(n+1)&=(n+1)\left({\frac {n}{2}}+1\right)\\&=(n+1)\left({\frac {n}{2}}+{\frac {2}{2}}\right)\\&=(n+1)\left({\frac {n+2}{2}}\right)\\&={\frac {(n+1)(n+2)}{2}}\\&={\frac {(n+1)((n+1)+1)}{2}}.\end{aligned}}}
