KVPY 2020-21 SA Exam Previous Papers

Question : 10 of 80

Marks: +1, -0

Let [x] be the greatest integer less than or equal to x, for a real number x. Then the following sum

[{2^{2020}+1}/{2^{2018}+1}]+[{3^{2020}+1}/{3^{2018}+1}]+[{4^{2020}+1}/{4^{2018}+1}]

+[{5^{2020}+1}/{5^{2018}+1}]+[{6^{2020}+1}/{6^{2018}+1}]

Solution:

{1+x^{2020}}/{1+x^{2018}}={x^{2}(1+x^{2018})+1-x^{2}}/{1+x^{2018}}

=x^{2}+{1-x^{2}}/{1+x^{2018}}

Put x = 2

[4+{(-3)}/{1+2^{2018}}]=3

Put x = 3

[9-{8}/{1+3^{2018}}]=8

Similarly for x = 4

[16-{15}/{1+4^{2018}}]=15

For x = 5

[25-{24}/{1+5^{2018}}]=24

For x = 6

[36-{35}/{1+6^{2018}}]=35

= 85

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