Iift 2008 Question 38
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IIFT 2008
The given series is 12 – 22 + 32 – 42 + ….. + 20012 – 20022 + 20032
12 – 22 can be written as = (1 + 2) (1 – 2) = 3 x (-1) = -3
32 – 42 can be written as = (3 +4) x (3 – 4) = 7 x (-1) = -7
52 – 62 can be written as = (5 + 6) x (5 – 6) = 11 x (-1) = -11
Therefore, all the terms till 20022 can be expressed as an AP.
The last term of the AP will be (2001 + 2002) (2001 – 2002) = -4003
Therefore, the given expression is reduced to -3 – 7 … - 4003 + 20032
Let is evaluate the value of -3 -7…. - 4003
Number of terms, n = 4003/4-3 + 1 = 1001
Sum n/2 = (first term + last term)
= 1001/2 x (-4006)
= -2005003
= 20032 = 4012009
Value of the given expression = 4012009 – 2005003 - 2007006
Therefore, option A is the right answer.