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1.
For all
a
>
0
n
+
a
>
n
n
>
n
−
a
n
>
n
−
a
1
>
n
−
a
n
1
n
−
a
>
1
n
{\displaystyle {\begin{matrix}a&>&0\\n+a&>&n\\n&>&n-a\\{\sqrt {n}}&>&{\sqrt {n-a}}\\1&>&{\frac {\sqrt {n-a}}{\sqrt {n}}}\\{\frac {1}{\sqrt {n-a}}}&>&{\frac {1}{\sqrt {n}}}\end{matrix}}}
Therefore
1
1
{\displaystyle {\frac {1}{\sqrt {1}}}}
1
2
{\displaystyle {\frac {1}{\sqrt {2}}}}
1
3
{\displaystyle {\frac {1}{\sqrt {3}}}}
>
1
n
{\displaystyle >{\frac {1}{\sqrt {n}}}}
When a>b and c>d, a+c>b+d ( See also Replace it if you find a better one).
Therefore we have:
1
1
+
1
2
+
1
3
.
.
.
.
.
.
+
1
n
>
n
×
1
n
{\displaystyle {\frac {1}{\sqrt {1}}}+{\frac {1}{\sqrt {2}}}+{\frac {1}{\sqrt {3}}}......+{\frac {1}{\sqrt {n}}}>n\times {\frac {1}{\sqrt {n}}}}
1
1
+
1
2
+
1
3
.
.
.
.
.
.
+
1
n
>
n
n
×
n
n
{\displaystyle {\frac {1}{\sqrt {1}}}+{\frac {1}{\sqrt {2}}}+{\frac {1}{\sqrt {3}}}......+{\frac {1}{\sqrt {n}}}>{\frac {n}{\sqrt {n}}}\times {\frac {\sqrt {n}}{\sqrt {n}}}}
1
1
+
1
2
+
1
3
.
.
.
.
.
.
+
1
n
>
n
n
n
{\displaystyle {\frac {1}{\sqrt {1}}}+{\frac {1}{\sqrt {2}}}+{\frac {1}{\sqrt {3}}}......+{\frac {1}{\sqrt {n}}}>{\frac {n{\sqrt {n}}}{n}}}
1
1
+
1
2
+
1
3
.
.
.
.
.
.
+
1
n
>
n
{\displaystyle {\frac {1}{\sqrt {1}}}+{\frac {1}{\sqrt {2}}}+{\frac {1}{\sqrt {3}}}......+{\frac {1}{\sqrt {n}}}>{\sqrt {n}}}
Let us call the proposition
(
n
0
)
+
(
n
1
)
+
(
n
2
)
+
.
.
.
+
(
n
n
)
=
2
n
{\displaystyle {n \choose 0}+{n \choose 1}+{n \choose 2}+...+{n \choose n}=2^{n}}
Assume this is true for some n, then
(
n
0
)
+
(
n
1
)
+
(
n
2
)
+
.
.
.
+
(
n
n
)
=
2
n
{\displaystyle {n \choose 0}+{n \choose 1}+{n \choose 2}+...+{n \choose n}=2^{n}}
2
×
{
(
n
0
)
+
(
n
1
)
+
(
n
2
)
+
.
.
.
+
(
n
2
)
}
=
2
n
+
1
{\displaystyle 2\times \left\{{n \choose 0}+{n \choose 1}+{n \choose 2}+...+{n \choose 2}\right\}=2^{n+1}}
{
(
n
0
)
+
(
n
n
)
}
+
{
(
n
0
)
+
2
(
n
1
)
+
2
(
n
2
)
+
.
.
.
+
2
(
n
n
−
1
)
+
(
n
n
)
}
=
2
n
+
1
{\displaystyle \left\{{n \choose 0}+{n \choose n}\right\}+\left\{{n \choose 0}+2{n \choose 1}+2{n \choose 2}+...+2{n \choose n-1}+{n \choose n}\right\}=2^{n+1}}
{
(
n
0
)
+
(
n
n
)
}
+
{
(
n
0
)
+
(
n
1
)
}
+
{
(
n
1
)
+
(
n
2
)
}
+
{
(
n
2
)
+
(
n
3
)
}
+
.
.
.
+
{
(
n
n
−
1
)
+
(
n
n
)
}
=
2
n
+
1
{\displaystyle \left\{{n \choose 0}+{n \choose n}\right\}+\left\{{n \choose 0}+{n \choose 1}\right\}+\left\{{n \choose 1}+{n \choose 2}\right\}+\left\{{n \choose 2}+{n \choose 3}\right\}+...+\left\{{n \choose n-1}+{n \choose n}\right\}=2^{n+1}}
Now using the identities of this function:
(
n
a
)
+
(
n
a
+
1
)
=
(
n
+
1
a
+
1
)
{\displaystyle {n \choose a}+{n \choose a+1}={n+1 \choose a+1}}
{
(
n
0
)
+
(
n
n
)
}
+
(
n
+
1
1
)
+
(
n
+
1
2
)
+
(
n
+
1
3
)
+
.
.
.
+
(
n
+
1
n
)
=
2
n
+
1
{\displaystyle \left\{{n \choose 0}+{n \choose n}\right\}+{n+1 \choose 1}+{n+1 \choose 2}+{n+1 \choose 3}+...+{n+1 \choose n}=2^{n+1}}
Since
(
n
0
)
=
(
n
n
)
=
1
{\displaystyle {n \choose 0}={n \choose n}=1}
(
n
+
1
0
)
+
(
n
+
1
n
+
1
)
+
(
n
+
1
1
)
+
(
n
+
1
2
)
+
(
n
+
1
3
)
+
.
.
.
+
(
n
+
1
n
)
=
2
n
+
1
{\displaystyle {n+1 \choose 0}+{n+1 \choose n+1}+{n+1 \choose 1}+{n+1 \choose 2}+{n+1 \choose 3}+...+{n+1 \choose n}=2^{n+1}}
(
n
+
1
0
)
+
(
n
+
1
1
)
+
(
n
+
1
2
)
+
(
n
+
1
3
)
+
.
.
.
+
(
n
+
1
n
)
+
(
n
+
1
n
+
1
)
=
2
n
+
1
{\displaystyle {n+1 \choose 0}+{n+1 \choose 1}+{n+1 \choose 2}+{n+1 \choose 3}+...+{n+1 \choose n}+{n+1 \choose n+1}=2^{n+1}}
Therefore P(n) implies P(n+1), and by simple substitution P(0) is true.
Therefore by the principal of mathematical induction, P(n) is true for all n. Alternate solution
(
a
+
b
)
n
=
(
n
0
)
a
n
+
(
n
1
)
a
n
−
1
b
+
⋯
+
(
n
n
)
b
n
{\displaystyle (a+b)^{n}={n \choose 0}a^{n}+{n \choose 1}a^{n-1}b+\cdots +{n \choose n}b^{n}}
letting a = b = 1, we get
(
1
+
1
)
n
=
2
n
=
(
n
0
)
+
(
n
1
)
+
⋯
+
(
n
n
)
{\displaystyle (1+1)^{n}=2^{n}={n \choose 0}+{n \choose 1}+\cdots +{n \choose n}}
as required.
5.
Let
P
(
x
)
=
x
n
+
y
n
{\displaystyle P(x)=x^{n}+y^{n}\,}
P
(
−
y
)
=
(
−
y
)
n
+
y
n
=
−
y
n
+
y
n
(
When n is an odd integer
)
=
0
{\displaystyle {\begin{matrix}P(-y)&=&(-y)^{n}+y^{n}\\\ &=&-y^{n}+y^{n}({\mbox{When n is an odd integer}})\\\ &=&0\end{matrix}}}
Therefore by factor theorem(link here please), (x-(-y))=(x+y) is a factor of P(x).
Since the other factor, which is also a polynomial, has integer value for all integer x,y and n (I've skipped the part about making sure all coeifficients are of integer value for this moment), it's now obvious that
x
n
+
y
n
x
+
y
{\displaystyle {\frac {x^{n}+y^{n}}{x+y}}}
