Dare2Solve | Infinite summation of a finite summation

This question was uploaded on 02/12/21 on social media accounts.

Solution:

\[f(x) = \frac1{2^x} + \frac1{3^x} + ..... + \frac1{4000^x} = \sum_{n=2}^{4000}\frac1{n^x}\]

Let sum of f(1), f(2), f(3), ....., is 'S'.

\[S = f(2)+f(3)+f(4)+..... = \sum_{x=2}^{\infty}\left(\sum_{n=2}^{4000}\frac1{n^x}\right)\]

\[S =  \sum_{x=2}^{\infty}\frac1{2^x} + \sum_{x=2}^{\infty}\frac1{3^x} + ....+\sum_{x=2}^{\infty}\frac1{4000^x}\]

Here, all terms are in infinite G.P.

\[\sum_{x=2}^{\infty}\frac1{a^x} = \frac{1/a^2}{1-1/a} = \frac1{a(a-1)} = \frac1{a-1} - \frac1{a}\]

\[\Rightarrow S = \left(\frac1{2-1} - \frac1{2}\right) + \left(\frac1{3-1} - \frac1{3}\right) + ....+\left(\frac1{4000-1} - \frac1{4000}\right)\]

\[\Rightarrow S = \left(\frac11 - \frac1{2}\right) + \left(\frac1{2} - \frac1{3}\right) + ....+\left(\frac1{3999} - \frac1{4000}\right)\]

It is a example of telescopic summation where only first and last term will remain and all other terms will cancel.

\[\Rightarrow S = 1 - \frac1{4000} = \frac{3999}{4000}\]

Dare2Solve | Sides of triangle are 1st, 3rd, 4th member of an increasing AP

This question was uploaded on 28/11/21 on social media accounts.

Solution:

By given conditions, all four terms are part of increasing A.P:

Let the first term be 'a' and the common difference is d:

\[a_1 = a\]

\[a_2 = a + d\]

\[a_3 = a + 2d\]

\[a_4 = a + 3d\]

Given that first term is smaller than last term:

\[\Rightarrow a_1 < a_2 < a_3 < a_4\]

Here 60 degrees angle is not at the opposite of the smallest or the largest side:

\[\Rightarrow A = 60^\circ\]

By using the cosine rule:

\[\cos60 = \frac{a^2+(a+2d)^2-(a+3d^2)}{2(a)(a+3d)}\]

\[\Rightarrow a=5d\]

By substituting these values:

\[a_1:a_3:a_4 = 5:7:8\]

Now by using the sine rule:

\[\frac7{\sin 60} = \frac8{\sin B} = \frac5{\sin C}\]

\[\color{red} { \Rightarrow \sin A = \frac{\sqrt3}{2}}\]

\[\color{red} { \Rightarrow \sin B = \frac{8\sqrt3}{14}}\]

\[\color{red} { \Rightarrow \sin C = \frac{5\sqrt3}{14}}\]

Dare2Solve | Cambridge university question in 1801 - Infinite summation

This question was uploaded on 26/11/21 on social media accounts.

This question was asked at Cambridge university in 1801.

Solution:

Given series:

\[S = \frac1{1\times3} - \frac1{2\times4} + \frac1{3\times5} - \frac1{4\times6} -.....\]

All terms are in the form of:

\[\frac1{(n)(n+2)}\]

\[\rightarrow \frac1{(n)(n+2)} = \frac12\left( \frac1n - \frac1{n+2} \right)\]

Now our given series:

\[S = \frac12\left[\left( \frac11 - \frac13 \right) - \left( \frac12 - \frac14 \right) + \left( \frac13 - \frac15 \right) - \left( \frac14 - \frac16 \right) + ....\right]\]

\[S = \frac12\left[1 - \frac12\right] = \frac14\]

Dare2solve | Infinite summation of converging series

This question was uploaded on 23/11/21 on social media accounts.

Solution:

Given Equation:

\[S = 1 + \frac12 + \frac1{2.4} + \frac1{2.4.6} + \frac1{2.4.6.8} + ....\]

\[S = 1 + \frac{\left(\frac12\right)}{1!} + \frac{\left(\frac12\right)^2}{2!} + \frac{\left(\frac12\right)^3}{3!} + \frac{\left(\frac12\right)^4}{4!} + ....\]

We know that:

\[e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ....\]

\[\Rightarrow S = e^{1/2} = \sqrt e\]

Dare2solve | Question from Putnam 1977 B1 - Continuous multiplication series

This question was uploaded on 05/11/21 on social media accounts.

This question was asked in Putnam 1977 B1 and I got this question from YouTube from a channel named "Letsthinkcritically"

Solution:

Let:

\[S = \frac{2^3-1}{2^3+1}\times\frac{3^3-1}{3^3+1}\times\frac{4^3-1}{4^3+1}\times\cdot\cdot\cdot\]

We know that:

\[a^3 - b^3 = (a - b) (a^2 + ab + b^2)\]

\[a^3 + b^3 = (a + b) (a^2 - ab + b^2)\]

By putting this in our original series, we will get:

\[S = \frac{1\times7}{3\times3}\times\frac{2\times13}{4\times7}\times\frac{3\times21}{5\times13}\times\cdot\cdot\cdot\]

\[\Rightarrow S = \left(\frac23\times\frac34\times\frac45\times\cdot\cdot\cdot\right)\left(\frac73\times\frac{13}7\times\frac{21}{13}\times\cdot\cdot\cdot\right)\]

\[\Rightarrow S = (2)\left(\frac13\right)=\frac23\]