If a1,…,an are in arithmetic progression and ai&gt;0 for all i, then how to prove the following two identities:1) 1a1+a2+1a2+a3+…+1an−1+an=n−1a1+an2) 1a1⋅an+1a2⋅an−1+1a3⋅an−2+…+1an⋅a1=2a1+an(1a1+1a2+…+1an)

For number 1:∑k=1n−11ak+ak+1=∑k=1n−1ak+1−akdwhere d is the common difference,=1d(an−a1)=an−a1d(an+a1)=n−1a1+an

For number 2:S=2∑k=1n1ak=(1a1+1an)+(1a2+1an−1)+…+(1an+1a1)=∑k=1nan−k+1+akakan−k+1Now an−k+1+ak=2a1+(n−1)d=a1+an and soS=(a1+an)∑k=1n1akan−k+1from which the result follows.

For number 1:∑k=1n−11ak+ak+1=∑k=1n−1ak+1−akdwhere d is the common difference,=1d(an−a1)=an−a1d(an+a1)=n−1a1+an

For number 2:S=2∑k=1n1ak=(1a1+1an)+(1a2+1an−1)+…+(1an+1a1)=∑k=1nan−k+1+akakan−k+1Now an−k+1+ak=2a1+(n−1)d=a1+an and soS=(a1+an)∑k=1n1akan−k+1from which the result follows.

Expert Answer to "If a1,…,an are in arithmetic progression and ai>0..."

Ari Sheppard

Answered question

2022-04-21

a_{1}, \dots, a_{n}

are in arithmetic progression and

a_{i} > 0

for all i, then how to prove the following two identities:
1)

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \dots + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}

\frac{1}{a_{1} \cdot a_{n}} + \frac{1}{a_{2} \cdot a_{n - 1}} + \frac{1}{a_{3} \cdot a_{n - 2}} + \dots + \frac{1}{a_{n} \cdot a_{1}} = \frac{2}{a_{1} + a_{n}} (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \dots + \frac{1}{a_{n}})

Answer & Explanation

Patricia Stanley

Beginner2022-04-22Added 10 answers

For number 1:

\sum_{k = 1}^{n - 1} \frac{1}{\sqrt{a_{k}} + \sqrt{a_{k + 1}}} = \sum_{k = 1}^{n - 1} \frac{\sqrt{a_{k + 1} - \sqrt{a_{k}}}}{d}

where d is the common difference,

= \frac{1}{d} (\sqrt{a_{n}} - \sqrt{a_{1}}) = \frac{a_{n} - a_{1}}{d (\sqrt{a_{n}} + \sqrt{a_{1}})}

= \frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}

Bethany Mills

Beginner2022-04-23Added 14 answers

For number 2:

S = 2 \sum_{k = 1}^{n} \frac{1}{a_{k}} = (\frac{1}{a_{1}} + \frac{1}{a_{n}}) + (\frac{1}{a_{2}} + \frac{1}{a_{n - 1}}) + \dots + (\frac{1}{a_{n}} + \frac{1}{a_{1}})

= \sum_{k = 1}^{n} \frac{a_{n - k + 1} + a_{k}}{a_{k} a_{n - k + 1}}

Now

a_{n - k + 1} + a_{k} = 2 a_{1} + (n - 1) d = a_{1} + a_{n}

and so

S = (a_{1} + a_{n}) \sum_{k = 1}^{n} \frac{1}{a_{k} a_{n - k + 1}}

from which the result follows.

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