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Derivative from First Principle

If a1, a2, ...

Question

If $a_{1}, a_{2}, a_{3}$ ,______ are in A.P. of common difference d, then prove that $S = {tan}^{- 1} \frac{d}{1 + a_{1} a_{2}} + {tan}^{- 1} \frac{d}{1 + a_{2} a_{3}} +$ _____ $+ {tan}^{- 1} \frac{d}{1 + a_{n} a_{n + 1}} = {tan}^{- 1} \frac{n d}{1 + a_{1} a_{n + 1}}$ .

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Solution

$d = a_{2} - a_{1} = a_{3} - a_{2} . . . = a_{n} - a_{n - 1} = a_{n + 1} - a_{n}$
$∴ T_{1} = {tan}^{- 1} a_{2} - {tan}^{- 1} a_{1}$
$T_{2} = {tan}^{- 1} a_{3} - {tan}^{- 1} a_{2}$
_______________________________________
$T_{n + 1} = {tan}^{- 1} a_{n + 1} - {tan}^{- 1} a_{n}$
$∴ S_{n} = {tan}^{- 1} a_{n + 1} - {tan}^{- 1} a_{1}$
$= {tan}^{- 1} \frac{a_{n + 1} - a_{1}}{1 + a_{1} a_{n + 1}} = {tan}^{- 1} \frac{n d}{1 + a_{1} a_{n + 1}}$ .

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Similar questions

a_{1}, a_{2}, a_{3} \dots \dots . . a_{n}

are in A.P. with common difference d, then

t a n [t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + t a n^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + \dots \dots \dots \dots \dots \dots + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =

θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}

a_{1}, a_{2}, a_{3}, \dots a_{n}

are in A.P. with common difference d, then

t a n θ =

θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}

a_{1}, a_{2}, a_{3}, \dots a_{n}

are in A.P. with common difference d, then

t a n θ =

a_{1}, a_{2}, \dots \dots a_{n + 1}

are in A.P. with common difference d, then

n \sum r = 1 t a n^{- 1} (\frac{d}{1 + a_{r} a_{r + 1}}) =

a_{1}, a_{2}, a_{3}, . . ., a_{n}

is an A.P. with common difference d, then

tan [{tan}^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + {tan}^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + . . . + {tan}^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] = \frac{(n - p) d}{q + a_{1} a_{n}}

.
Find the value of

p + q

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