Question

1 – 2 + 3 - 4 - 5 - 6+ ...
+ 2009 - 2010 + 2011 - 2012.(NSTSE 2013)
(A) - 2000
(B) - 1
(C) 1000
(D) -10062
(D) - 1006

​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{1-2+3-4+5-\;.\;.\;.\;.-2010+2011-2012}$

$\underline{\textbf{To find:}}$

$\textsf{The sum}$

$\mathsf{1-2+3-4+5-\;.\;.\;.\;.-2010+2011-2012}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{1-2+3-4+5-\;.\;.\;.\;.-2010+2011-2012}$

$\mathsf{=(1+3+\;.\;.\;.\;.+2011)-(2+4+6\;.\;.\;.\;.+2012)}$

$\mathsf{=S_1-S_2}$

$\underline{\mathsf{S_1:}}$

$\mathsf{1+3+\;.\;.\;.\;.+2011}$

$\textsf{This is an A.P series with a=1, d=2}$

$\mathsf{n=\dfrac{l-a}{d}+1}$

$\mathsf{n=\dfrac{2011-1}{2}+1}$

$\mathsf{n=\dfrac{2010}{2}+1}$

$\mathsf{n=1005+1}$

$\mathsf{n=1006}$

$\mathsf{S_1=\dfrac{n}{2}[a+l]}$

$\mathsf{S_1=\dfrac{1006}{2}[1+2011]}$

$\mathsf{S_1=\dfrac{1006}{2}[2012]}$

$\mathsf{S_1=1006{\times}1006}$

$\underline{\mathsf{S_2:}}$

$\mathsf{2+4+6+\;.\;.\;.\;.\;.+2012}$

$\mathsf{=2(1+2+3+\;.\;.\;.\;.\;.+1006)}$

$\mathsf{=2{\times}\dfrac{n(n+1)}{2}\;where\;n=1006}$

$\mathsf{=1006{\times}1007}$

$\underline{\mathsf{Required\;sum:}}$

$\mathsf{=S_1-S_2}$

$\mathsf{=1006{\times}1006-1006{\times}1007}$

$\mathsf{=1006{\times}(1006-1007)}$

$\mathsf{=1006{\times}(-1)}$

$\mathsf{=-1006}$

$\underline{\textbf{Answer:}}$

$\mathsf{Option\;(D)\;is\;correct}$

$\underline{\textbf{Find more:}}$

Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

$\sf{1 - 2 + 3 - 4 + 5 - 6 + ... + 2009 - 2010 + 2011 - 2012}$

(A) - 2000

(B) - 1

(C) 1000

(D) - 1006

EVALUATION

Here the given expression is

$\sf{1 - 2 + 3 - 4 + 5 - 6 + ... + 2009 - 2010 + 2011 - 2012}$

The number of terms = 2012

We group them using brackets as below

$\sf{(1 - 2 )+ (3 - 4) + (5 - 6) + ... + (2009 - 2010 )+ (2011 - 2012)}$

Since each bracket contains two Terms

So total number of brackets = 1006

On simplification we get

$\sf{1 - 2 + 3 - 4 + 5 - 6 + ... + 2009 - 2010 + 2011 - 2012}$

$\sf{ = (1 - 2 )+ (3 - 4) + (5 - 6) + ... + (2009 - 2010 )+ (2011 - 2012)}$

$\sf{ = ( - 1)+ ( - 1) + ( - 1) + ... + ( - 1 )+ ( - 1)}$

Since on simplification each bracket we get the value - 1 and total number of brackets = 1006

So we get

$\sf{1 - 2 + 3 - 4 + 5 - 6 + ... + 2009 - 2010 + 2011 - 2012}$

$\sf{ = ( - 1)+ ( - 1) + ( - 1) + ... + ( - 1 )+ ( - 1)}$

$\sf{ = ( - 1) \times 1006}$

$\sf{ = - 1006}$

FINAL ANSWER

Hence the correct option is (D) - 1006

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