show that 1 upon 1 + root 2 + 1 upon root 2 + root 3 is equals to minus 1 root 3
Answers
Step-by-step explanation:
this is ur answer
see the photo
Answer:
Given:
\frac{1}{1+\sqrt 2}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{8}+\sqrt{9}}
1+
2
1
+
2
+
3
1
+....+
8
+
9
1
To find:
Value of \frac{1}{1+\sqrt 2}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{8}+\sqrt{9}}
1+
2
1
+
2
+
3
1
+....+
8
+
9
1
Solution:
By rationalization
\frac{1}{1+\sqrt 2}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}+\frac{1}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+....+\frac{1}{\sqrt{8}+\sqrt{9}}\times \frac{\sqrt{9}-\sqrt{8}}{\sqrt{9}-\sqrt{8}}
1+
2
1
×
2
−1
2
−1
+
2
+
3
1
×
3
−
2
3
−
2
+....+
8
+
9
1
×
9
−
8
9
−
8
\frac{\sqrt{2}-1}{(\sqrt{2})^2-1}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4})^2-(\sqrt{3})^2}+....\frac{\sqrt{9}-\sqrt{8}}{(\sqrt{9})^2-(\sqrt{8})^2}
(
2
)
2
−1
2
−1
+
(
3
)
2
−(
2
)
2
3
−
2
+
(
4
)
2
−(
3
)
2
4
−
3
+....
(
9
)
2
−(
8
)
2
9
−
8
Using identity:
(a+b)(a-b)=a^2-b^2(a+b)(a−b)=a
2
−b
2
\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+\frac{\sqrt{5}-\sqrt{4}}{5-4}+\frac{\sqrt{6}-\sqrt{5}}{6-5}+\frac{\sqrt{7}-\sqrt{6}}{7-6}+\frac{\sqrt{8}-\sqrt{7}}{8-7}+\frac{\sqrt{9}-\sqrt{8}}{9-8}
2−1
2
−1
+
3−2
3
−
2
+
4−3
4
−
3
+
5−4
5
−
4
+
6−5
6
−
5
+
7−6
7
−
6
+
8−7
8
−
7
+
9−8
9
−
8
\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+\sqrt{7}-\sqrt{6}+\sqrt{8}-\sqrt{7}+\sqrt{9}-\sqrt{8}
2
−1+
3
−
2
+
4
−
3
+
5
−
4
+
6
−
5
+
7
−
6
+
8
−
7
+
9
−
8
=-1+3=−1+3
=2=2
\frac{1}{1+\sqrt 2}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{8}+\sqrt{9}}=2
1+
2
1
+
2
+
3
1
+....+
8
+
9
1
=2
Step-by-step explanation:
done