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STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
$\left[\frac{2^{2020}+1}{2^{2018}+1}\right]+\left[\frac{3^{2020}+1}{3^{2018}+1}\right]+\left[\frac{4^{2020}+1}{4^{2018}+1}\right] +\left[\frac{5^{2020}+1}{5^{2018}+1}\right] + \left[\frac{6^{2020}+1}{6^{2018}+1}\right]$  is
  • A
    $80$
  • ✓
    $85$
  • C
    $90$
  • D
    $95$
✓

Answer

Correct option: B.
$85$
b
(b)

$\frac{1+x^{2020}}{1+x^{2018}}=\frac{x^2\left(1+x^{2018}\right)+1-x^2}{1+x^{2018}}$

$=x^2+\frac{1-x^2}{1+x^{2018}}$

Put $x=2 \therefore\left[4+\frac{(-3)}{1+2^{2018}}\right]=3$

Put $x=3 \therefore\left[9-\frac{8}{1+3^{2018}}\right]=8$

Similarly for $x=4,\left[16-\frac{15}{1+4^{2018}}\right]=15$

For $x=5,\left[25-\frac{24}{1+5^{2018}}\right]=24$

For $x=6,\left[36-\frac{35}{1+6^{2018}}\right]=35$

$\therefore$ Required sum

$=3+8+15+24+35=85$

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