The value of left[frac{2^{2020}+1}{2^{2018}+1 | vedclass

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Question

(B) Let $f(x) = \frac{x^{2020}+1}{x^{2018}+1}$.We can rewrite this as $f(x) = \frac{x^2(x^{2018}+1) + 1 - x^2}{x^{2018}+1} = x^2 + \frac{1-x^2}{x^{201

Vedclass · Accepted Answer

$85$

Vedclass · Answer

(B) Let $f(x) = \frac{x^{2020}+1}{x^{2018}+1}$.We can rewrite this as $f(x) = \frac{x^2(x^{2018}+1) + 1 - x^2}{x^{2018}+1} = x^2 + \frac{1-x^2}{x^{2018}+1}$.For $x \ge 2$,the term $\frac{1-x^2}{x^{2018}+1}$ is negative and very close to $0$ (specifically,it lies between $-1$ and $0$).Therefore,$[f(x)] = [x^2 - \epsilon] = x^2 - 1$ for $x \ge 2$,where $0 Applying this to each term:For $x=2: [f(2)] = 2^2 - 1 = 3$.For $x=3: [f(3)] = 3^2 - 1 = 8$.For $x=4: [f(4)] = 4^2 - 1 = 15$.For $x=5: [f(5)] = 5^2 - 1 = 24$.For $x=6: [f(6)] = 6^2 - 1 = 35$.Summing these values: $3 + 8 + 15 + 24 + 35 = 85$.

The value of $\left[\frac{2^{2020}+1}{2^{2018}+1}\right]+\left[\frac{3^{2020}+1}{3^{2018}+1}\right]+\left[\frac{4^{2020}+1}{4^{2018}+1}\right] +\left[\frac{5^{2020}+1}{5^{2018}+1}\right] + \left[\frac{6^{2020}+1}{6^{2018}+1}\right]$ is (where $[\cdot]$ denotes the greatest integer function):

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