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Quadratic Equation and Inequalities question

2021 · 25 Jul · Shift 2 · Q32
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  5. /2021 · 25 Jul · Shift 2 · Q32

Quadratic Equation and Inequalities question

2021 · 25 Jul · Shift 2 · Q32

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If [x] be the greatest integer less than or equal to x, then ∑n=8100[(−1)nn2]\sum\limits_{n = 8}^{100} {\left[ {{{{{( - 1)}^n}n} \over 2}} \right]}n=8∑100​[2(−1)nn​] is equal to :
  1. A
    0
  2. B
    4
  3. C
    −-− 2
  4. D
    2
View written solutionFree

Correct answer: B

  1. We need to evaluate S=∑n=8100[(−1)nn2],S=\sum_{n=8}^{100}\left[\frac{(-1)^n n}{2}\right],S=∑n=8100​[2(−1)nn​], where [x][x][x] denotes the greatest integer less than or equal to xxx.

  2. Split according to parity of nnn.

  • If nnn is even, then (−1)n=1(-1)^n=1(−1)n=1, so [(−1)nn2]=[n2]=n2\left[\frac{(-1)^n n}{2}\right]=\left[\frac{n}{2}\right]=\frac{n}{2}[2(−1)nn​]=[2n​]=2n​ since n/2n/2n/2 is already an integer.

  • If nnn is odd, then (−1)n=−1(-1)^n=-1(−1)n=−1, so [(−1)nn2]=[−n2].\left[\frac{(-1)^n n}{2}\right]=\left[-\frac{n}{2}\right].[2(−1)nn​]=[−2n​]. For odd n=2k+1n=2k+1n=2k+1, −n2=−2k+12=−k−12,-\frac{n}{2}=-\frac{2k+1}{2}=-k-\frac12,−2n​=−22k+1​=−k−21​, hence [−n2]=−k−1=−n+12.\left[-\frac{n}{2}\right]=-k-1=-\frac{n+1}{2}.[−2n​]=−k−1=−2n+1​.

  1. Pair consecutive terms: one even and the next odd.

Let even n=2kn=2kn=2k, then the next odd term is 2k+12k+12k+1. Their contribution is [2k2]+[−2k+12]=k+(−k−1)=−1.\left[\frac{2k}{2}\right]+\left[-\frac{2k+1}{2}\right]=k+(-k-1)=-1.[22k​]+[−22k+1​]=k+(−k−1)=−1.

  1. Now count such pairs from n=8n=8n=8 to n=99n=99n=99: (8,9),(10,11),…,(98,99).(8,9),(10,11),\dots,(98,99).(8,9),(10,11),…,(98,99). This is an arithmetic sequence of even starts from 888 to 989898. Number of pairs is 98−82+1=45+1=46.\frac{98-8}{2}+1=45+1=46.298−8​+1=45+1=46. So sum up to 999999 is 46×(−1)=−46.46\times(-1)=-46.46×(−1)=−46.

  2. The last remaining term is n=100n=100n=100: [(−1)1001002]=[50]=50.\left[\frac{(-1)^{100}100}{2}\right]=\left[50\right]=50.[2(−1)100100​]=[50]=50.

  3. Therefore, S=−46+50=4.S=-46+50=4.S=−46+50=4.

  4. So the correct option is 4.\boxed{4}.4​.

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