Topic : Sequence- Sum
from: category_eng
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What is the difference between the sum of the first 2003 even counting numbers and the sum of the first 2003 odd counting numbers?

mathrm{(A) } 0qquad mathrm{(B) } 1qquad mathrm{(C) } 2qquad mathrm{(D) } 2003qquad mathrm{(E) } 4006

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3.
Manipulation
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Manipulation
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Solution 1


The first 2003 even counting numbers are 2,4,6,...,4006.

The first 2003 odd counting numbers are 1,3,5,...,4005.

Thus, the problem is asking for the value of (2+4+6+...+4006)-(1+3+5+...+4005).

(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)

= 1+1+1+...+1 = oxed{mathrm{(D)} 2003}


Solution 2


Using the sum of an arithmetic progression formula, we can write this as frac{2003}{2}(2 + 4006) - frac{2003}{2}(1 + 4005) = frac{2003}{2} cdot 2 = oxed{mathrm{(D)} 2003}.




Solution 3


The formula for the sum of the first n even numbers, is S_E=n^{2}+n, (E standing for even).

Sum of first n odd numbers, is S_O=n^{2}, (O standing for odd).

Knowing this, plug 2003 for n,

S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 Rightarrow oxed{mathrm{(D)} 2003}.
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Calculation

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Complex