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User:Michael Hardy
Algebra will now establish that
n
(
n
+
1
)
2
+
(
n
+
1
)
=
(
n
+
1
)
(
(
n
+
1
)
+
1
)
2
,
{\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.)
n
(
n
+
1
)
2
+
(
n
+
1
)
=
(
n
+
1
)
(
n
2
+
1
)
=
(
n
+
1
)
(
n
2
+
2
2
)
=
(
n
+
1
)
(
n
+
2
2
)
=
(
n
+
1
)
(
n
+
2
)
2
=
(
n
+
1
)
(
(
n
+
1
)
+
1
)
2
.
{\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)
holds.