can you solve every thing except 6,7

Solution 14: Let be a sequence of real numbers defined by
Solution 14: Let
1
{}
nn
x
³
be a sequence of real numbers defined by
1
,
xa
=


2
,
xb
=
22
1
()
2
nnn
xxx
++
=+
for n =1, 2…
We know,
211
211
211
2
1
();
2
1
();
2
1
;
2
nnnn
nnnn
nnnn
xxxx
xxxx
xxxx
+++
+++
++-
-=-
-=-
-=-
Putting down the above given values of
1
,
xa
=

2
,
xb
=
22
1
()
2
nnn
xxx
++
=+
for n =1, 2… we get,
2121
21
1
();
2
1
();
2
nn
n
nn
n
xxxx
xxba
++
++
-=-
-=-
If n > m,
11221
231
2
...;
111
...
222
1
2
nmnnnnnmm
nm
nnm
m
xxxxxxxxx
xxba
ba
----+
---
-
-=-+-++++
ìü
-£+++-
íý
îþ
£-
For this reason,
1
{}
nn
x
³
is a Cauchy series.
Solution 15: For the metric space to be (X, d) where,
1
(,)
xy
dxy
xy
ì-+
ï
=
í
-
ï
î
Where x, y or both is positive.
The point is that now using the assumptions, that (xn) converges to x and (xn) converges
to y,
(,)(,)(,)
nn
dxydxxdxy
£+
Suppose that (X, d) is a metric space and that the sequence (xn) in X converges to x and y.
Let
0
Î>
Since (xn) converges to x, there exists an index such that:
(,)1
n
dxxxy
<-+
and
(,)
n
dxyxy
<-
Let N = max {N1; N2}. Then both inequalities above hold, and so the triangle inequality gives:
(,)(,)(,)
nn
dxydxxdxy
£+
Or,
(,)(,)(,)
nn
dxydxxdxy
£+
;
(,)1
dxyxyxy
£-++-
;
This implies that d(x; y) > 0, which in turn implies that x = y.
We conclude that the metric space of a convergent sequence is unique.
For the metric space to be (X, d) where,...