Question

If $M=[1+ \frac{1}{2^\frac{2}{3}}+frac{1}{3^\frac{2}{3}}+ ...+frac{1}{1000^\frac{2}{3}}]$ , where [.] denotes greatest integer function, then find the value of $M - 20$​

Answer 1

Answer:

The solution set of ln(5−7x)⩽1 is given by,

A. [5−e7,57)

B. [2−e3,23)

C. (−10,7)

D. All real numbers

Answer 2

We can approximate the given series as an integral using the following inequality:

$\rm \frac{1}{(k+1)^{\frac{2}{3}}} < \int_{k}^{k+1} \frac{1}{x^{\frac{2}{3}}} dx < \frac{1}{k^{\frac{2}{3}}}$

Integrating both sides of the inequality from 1 to 1000, we get:

$\rm \int_{1}^{1001} \frac{1}{x^{\frac{2}{3}}} dx < M < 1 + \int_{1}^{1000} \frac{1}{x^{\frac{2}{3}}} dx$

Evaluating the integrals, we get:

$\rm \frac{3}{2} (1 - \frac{1}{1001^{\frac{1}{3}}}) < M < 1 + \frac{3}{2}(1 - \frac{1}{1000^{\frac{1}{3}}})$

Using this inequality, we can find the greatest integer function of M as:

$\rm \lfloor M \rfloor = 1 + \lfloor \frac{3}{2}(1 - \frac{1}{1000^{\frac{1}{3}}}) \rfloor$

Evaluating this expression, we get:

$\rm \lfloor M \rfloor = 1$

Therefore, $\rm M - 20 = (1 + \frac{3}{2}(1 - \frac{1}{1001^{\frac{1}{3}}})) - 20 \\ = \frac{1}{2}(41 - \frac{3}{1001^{\frac{1}{3}}})$

Using a calculator, we can evaluate $\rm M - 20$ as:

$\rm M - 20 \approx 20.998$

Note that this is an approximation since we used an inequality to estimate the value of M. The actual value of M could be slightly different, but it should be very close to this approximation.