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If \(x = 1 + 2 + 3 + ... + 100\), and \(y = 1 + 3 + 5 + ... + 199\), then what is the value of y-x?
A. 4850
B. 4900
C. 4950
D. 5000
E. 5050
A. 4850
B. 4900
C. 4950
D. 5000
E. 5050
24 Oct 2022, 19:27
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If \(x = 1 + 2 + 3 + ... + 100\), and \(y = 1 + 3 + 5 + ... + 199\), then what is the value of y-x?
A. 4850
B. 4900
C. 4950
D. 5000
E. 5050
A. 4850
B. 4900
C. 4950
D. 5000
E. 5050
Use formulas:
Sum of first n positive integers = \(\frac{n(n+1)}{2}\)
So, sum of first 100 positive integers = \(\frac{100(100+1)}{2}=5050\)
Sum of first n positive odd integers = \(n^2\)
So, sum of first 100 positive odd integers = \(n^2=100^2=10000\)
y-x=10000-5050=4950
Second way:
\(y = 1 + 3 + 5 + ... + 199\)
\(y +100= 1 + 3 + 5 + ... + 199+100\)
\(y +100= 1 +1+ 3 +1+ 5 +1+ ... + 199+1\)
\(y+100=2+4+6+…..200=2(1+2+3+…+100)\)
\(y+100-x=2(1+2+3+…+100)-(1+2+3+….+100)=1+2+3+…..100=100*101/2=5050\)
\(y-x=5050-100=4950\)
C
