All-the-terms-of-the-arithmetic-progession-u-1-u-2-u-3-u-n-are-positive-use-induction-to-prove-that-for-n-2-1-u-1-u-2-1-u-2-u-3-1-u-3-u-4-1-u-n-1-u-n-

Question Number 7743 by Tawakalitu. last updated on 13/Sep/16

A l l t h e t e r m s o f t h e a r i t h m e t i c p r o g e s s i o n

u_{1}, u_{2}, u_{3}, \dots u_{n} a r e p o s i t i v e . u s e i n d u c t i o n t o

p r o v e t h a t f o r n ⩾ 2

\frac{1}{u_{1} u_{2}} + \frac{1}{u_{2} u_{3}} + \frac{1}{u_{3} u_{4}} + \dots \frac{1}{u_{n - 1} u_{n}} = \frac{n - 1}{u_{1} u_{n}}

Answered by Yozzia last updated on 13/Sep/16

L e t P (n) d e n o t e t h e s t a t e m e n t, \forall n \in N, n ⩾ 2,

\frac{1}{u_{1} u_{2}} + \frac{1}{u_{2} u_{3}} + \frac{1}{u_{3} u_{4}} + \dots + \frac{1}{u_{n - 1} u_{n}} = \frac{n - 1}{u_{1} u_{n}}

f o r u_{1}, u_{2}, \dots, u_{n} b e i n g a n A . P w i t h a l l t e r m s p o s i t i v e .

F o r n = 2, P (2) s a y s \frac{1}{u_{1} u_{2}} = \frac{2 - 1}{u_{1} u_{2}} = \frac{1}{u_{1} u_{2}}

w h i c h i s t r u e .

A s s u m e P (n) i s t r u e f o r s o m e n = k, i . e

\underset{i = 1}{\sum^{k - 1}} \frac{1}{u_{i} u_{i + 1}} = \frac{k - 1}{u_{1} u_{k}} .

\underset{i = 1}{\sum^{k - 1}} \frac{1}{u_{i} u_{i + 1}} + \frac{1}{u_{k} u_{k + 1}} = \frac{k - 1}{u_{1} u_{k}} + \frac{1}{u_{k} u_{k + 1}}

\underset{i = 1}{\sum^{k}} \frac{1}{u_{i} u_{i + 1}} = \frac{1}{u_{k}} (\frac{k - 1}{u_{1}} + \frac{1}{u_{k + 1}})

\underset{i = 1}{\sum^{k}} \frac{1}{u_{i} u_{i + 1}} = \frac{1}{u_{k}} (\frac{(k - 1) u_{k + 1} + u_{1}}{u_{1} u_{k + 1}})

F o r t h e A . P, u_{t} = u_{1} + (t - 1) d f o r t \in N, d = c o m m o n d i f f e r e n c e .

∴ u_{k + 1} = u_{1} + k d

\underset{i = 1}{\sum^{k}} \frac{1}{u_{i} u_{i + 1}} = \frac{1}{u_{k}} (\frac{(k - 1) (u_{1} + k d) + u_{1}}{u_{1} u_{k + 1}})

\underset{i = 1}{\sum^{k}} \frac{1}{u_{i} u_{i + 1}} = \frac{1}{u_{k}} (\frac{k (u_{1} + k d) - u_{1} - k d + u_{1}}{u_{1} u_{k + 1}})

\underset{i = 1}{\sum^{k}} \frac{1}{u_{i} u_{i + 1}} = \frac{1}{u_{k}} (\frac{k (u_{1} + k d - d)}{u_{1} u_{k + 1}}) = \frac{1}{u_{k}} \times \frac{k (u_{1} + (k - 1) d)}{u_{1} u_{k + 1}}

\underset{i = 1}{\sum^{k}} \frac{1}{u_{i} u_{i + 1}} = \frac{1}{u_{k}} \times \frac{{k u}_{k}}{u_{1} u_{k + 1}} = \frac{k}{u_{1} u_{k + 1}} w h i c h i s P (k + 1)

T h e r e f o r e, i f P (k) i s t r u e, t h e n P (k + 1) i s t r u e .

S i n c e P (2) i s t r u e t h e n, b y P . M . I,

P (n) i s t r u e \forall n \in N, n ⩾ 2 .

Commented by Tawakalitu. last updated on 13/Sep/16

W o w, T h a n k y o u s o m u c h .

Commented by Tawakalitu. last updated on 13/Sep/16

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