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by dtweah » Thu Apr 23, 2009 4:24 am
What is the value of the largest term in a sequence of consecutive odd integers beginning with 1 whose sum is 2^2024?

A. -1 + 2^1011

B. -1+ 2^1012

C. -1 + 2^1013

D. -1 + 2^2024

E. -1 + 2^2000
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by scoobydooby » Thu Apr 23, 2009 4:47 am
the sequence: (1, 3, 5, 7........)
sum to n terms: n/2{2a+(n-1)d}=2^2024 {AP with a=1, d=2}

=>n/2{2*1+(n-1)2}=2^2024
=>n*2n/2=2^2024
=>n^2=2^2024
=>n=2^(2024/2)
=>n=2^1012

let the last term be Tn.
Tn=a+(n-1)d
=>1+(2^1012-1)*2
=>1+2*2^1012-2
=-1+2*2^1012
=-1+2^1013

hence, C
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by dtweah » Thu Apr 23, 2009 1:30 pm
scoobydooby wrote:the sequence: (1, 3, 5, 7........)
sum to n terms: n/2{2a+(n-1)d}=2^2024 {AP with a=1, d=2}

=>n/2{2*1+(n-1)2}=2^2024
=>n*2n/2=2^2024
=>n^2=2^2024
=>n=2^(2024/2)
=>n=2^1012

let the last term be Tn.
Tn=a+(n-1)d
=>1+(2^1012-1)*2
=>1+2*2^1012-2
=-1+2*2^1012
=-1+2^1013

hence, C
1
1+3= 2^2
1+3+ 5= 3^2
1+ 3+5+7=4^2
1+ 3+5+7+9=5^2
So n^2 can always be expressed as the sum of the first consecutive positive odd integers beginning with 1. In fact the sum of the first n consecutive positive odd integers is n^2. We known the nth term for an odd integer is 2M-1.
2^2024=2^(1012)^2. Let M=2^(1012). Then we have
1 +3 +5 +7 +.... + 2^(1012)^2.

Then the last odd integer before the last term above is: 2 x 2^(1012)-1= -1+2^1013.
Choose C.
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