Question bank/JEE Main/Mathematics
JEE MainMathematicsModerateSequence & Series
2 × 10^8 1 2
3 × 2^8 - n -.
Show answer and solution
one digit nos. 2^1
one digit nos. 2^2
eight digit nos. 2^8
nine digit nos. 2^8
total = 2^1 + 2^2 +...+ 2^8 + 2^8
⇒n = 2^9 - 2 + 28
⇒n = 3 ⋅ 2^8 - 2
⇒2 = 3 ⋅ 2^8 - n
one digit nos. 2^2
eight digit nos. 2^8
nine digit nos. 2^8
total = 2^1 + 2^2 +...+ 2^8 + 2^8
⇒n = 2^9 - 2 + 28
⇒n = 3 ⋅ 2^8 - 2
⇒2 = 3 ⋅ 2^8 - n
Source: Grass-G / M-04 Paper 2 Mathematics 18-8-2020_Final.docx
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