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JEE Main 202311 AprilMorning Shift
Exam
JEE Main
Year
2023
Date
11 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2023 · 11 April · Morning Shift · Mathematics · Sequences and Series

Let S = 109 + (108)/(5) + 107 5 ^ 2 + . + 2 5 ^ 107 + 1 5 ^ 108 ....

Let S=109+1085+10752+….+25107+15108S = 109 + \dfrac{108}{5} + \dfrac{107}{{5}^{2}} + \ldots . + \dfrac{2}{{5}^{107}} + \dfrac{1}{{5}^{108}}. Then the value of (16S−(25)−54)\left(16S - ( 25{)}^{- 54}\right) is equal to

Correct answer

0

Step-by-step solution

Given,

S=109+1085+10752+…+25107+15108....(1)S = 109 + \dfrac{108}{5} + \dfrac{107}{{5}^{2}} + \ldots + \dfrac{2}{{5}^{107}} + \dfrac{1}{{5}^{108}} . . . . \left(1\right)

S5=1095+10852+…+…+25108+15109....(2)\dfrac{S}{5} = \dfrac{109}{5} + \dfrac{108}{{5}^{2}} + \ldots + \ldots + \dfrac{2}{{5}^{108}} + \dfrac{1}{{5}^{109}} . . . . \left(2\right)

Now subtracting above equations we get,
−4S5=−109+(15+152+…+15108+15109)\dfrac{- 4S}{5} = - 109 + \left(\dfrac{1}{5} + \dfrac{1}{{5}^{2}} + \ldots + \dfrac{1}{{5}^{108}} + \dfrac{1}{{5}^{109}}\right)

⇒−4S5=−109+15(1−(15)109)1−15⇒ \dfrac{- 4S}{5} = - 109 + \dfrac{1}{5}\dfrac{\left(1 - {\left(\dfrac{1}{5}\right)}^{109}\right)}{1 - \dfrac{1}{5}}

⇒4S5=−109+14(1−15109)⇒ \dfrac{4S}{5} = - 109 + \dfrac{1}{4}\left(1 - \dfrac{1}{{5}^{109}}\right)

⇒16S5=−436+(1−15109)⇒ \dfrac{16S}{5} = - 436 + \left(1 - \dfrac{1}{{5}^{109}}\right)

⇒16S=−436×5+(5−5−108)⇒ 16S = - 436 \times 5 + \left(5 - {5}^{- 108}\right)

⇒16S−(25)−54=2175⇒ 16S - ( 25{)}^{- 54} = 2175

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