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ariesvang

  • 3 years ago

Evaluate the series 1 + 4 + 16 + 64 + 256 + 1024.

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  1. ariesvang
    • 3 years ago
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    howd yuu get 2^11 - 1 = 2047?

  2. simamura
    • 3 years ago
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    This is geometric progression with first member a1=1 and q=4, so sum is 1*(1-4^6)/(1-4)=1365

  3. FoolForMath
    • 3 years ago
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    In general, \[ \sum \limits_ {i=0} ^n 2^i = 2^{n+1}-1\]

  4. simamura
    • 3 years ago
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    @FoolForMath this formula doesn't work here because there is no 8 (2^3) in sum

  5. sahil51993
    • 3 years ago
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    Further taking simamura's solution 4^6-1=4^2-1(4^4+1+4^2) =15(273) divided by 3 ie 1365

  6. ariesvang
    • 3 years ago
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    thank you guys for the help i appreciate it

  7. FoolForMath
    • 3 years ago
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    OKay I can see it now, 1 + 4 + 16 + 64 + 256 + 1024. This is a geometric series with first term 1, and common difference 4.

  8. sheg
    • 3 years ago
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    \[\large{1^2+2^2+4^2+8^2+16^2+32^2}\] now you can solve

  9. Zarkon
    • 3 years ago
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    there are only 6 numbers...just add them

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