2020/011) Prove that $4^{79}< 2^{100} + 3^{100} < 4^{80}$

We need to prove 2 inequalities
$4^{79}< 2^{100} + 3^{100} $ and $2^{100} + 3^{100} < 4^{80}$

For

$4^{79}< 2^{100} + 3^{100} $

If we prove that  $4^{79} < 3^{100} $ we are through

We have $\frac{4^{80}}{3^{100}} = (\frac{4^4}{3^5})^{20} = (\frac{256}{243})^{20}\cdots(1)$

Now $\frac{256}{243} = 1+\frac{13}{243} < 1 + \frac{13}{13 *18} = 1 + \frac{1}{18} = \frac{19}{18}\cdots(2)$

Hence $(\frac{256}{243})^{20} < (\frac{19}{18})^{20} = (\frac{361}{324})^{10}=  (1 + \frac{37}{324})^{10}\cdots(2)$

We have
$\frac{37}{324} < \frac{37}{37 * 8} < \frac{1}{8}$

From (2) and above
$(\frac{256}{243})^{20} < (1 + \frac{1}{8})^{10} = (\frac{9}{8})^{10} = (\frac{81}{64})^5 < (\frac{81}{63})^5 = (\frac{9}{7})^5 = \frac{59049}{16807} < 4$

From (1) and above $\frac{4^{80}}{3^{100}} < 4 $ and hence $4^{79} < 3^{100}$

Hence $4^{79} < + 2^{100} + 3^{100}$

For proving the $2^{nd}$ part we have

  $(\frac{256}{243})^20 = (1+  \frac{13}{243})^{20} > 1+  \frac{13}{243} * 20  > 1+ \frac{260}{243} > 2$ by bionomial expansion and deleting positive terms

Form above and (1) we have

$\frac{4^{80}}{3^{100}} > 2$ or $4^{80} > 2 * 3^{100} > 3^{100} + 2^{100} $

Proved

2020/010) If the binomial expansion of (a-b)n, n≥5, the sum of the 5th and 6th terms is zero.

The $\frac{a}{b}$ equals  a. $\frac{n-5}{6}$ b. $\frac{n-4}{5}$ c. $\frac{5}{n-4}$ d. $\frac{6}{n-5}$ 

Solution:
we have

$5^{th}$ term = ${n \choose 4} a^{n-4} (-b)^4$
$6^{th}$ term = ${n \choose 5} a^{n-5} (-b)^5$

As sum of the two is zero we have ${n \choose 4} a^{n-4} (-b)^4 + {n \choose 5} a^{n-5} (-b)^5 = 0$

or ${n \choose 4} a - {n \choose 5} (b) = 0$

or $\frac{a}{b} = \frac{n \choose 5}{n \choose 4}$

$= \frac{\frac{n!}{5!(n-5)!}}{\frac{n!}{4!(n-4)!}}$

$= \frac{(n-4)!}{(n-5)! 5}= \frac{n-4}{5}$

Hence ans (b)

2020/009) Solve in real $(x+4)(x+5)(x+6)(x+7) = 1680$

We have as  4+7 = 5 + 6

$((x+4)(x+7))((x+5)(x+6)) = 1680$
Or $x^2+11x+28)(x^2+11x+30) = 1680$
Putting $x^2+11x + 29= t$ we get
$(t-1)(t+1) = 1680$
Or $t^2 -1 = 1680$ or $t^2 = 1681$ or $t=\pm 41$

So we get 2 equations $x^2+11x+29=41$ or $x^2+11x+29=-41$

1st equation gives $x^2+11x-12=0$ or $(x+12)(x-1) = 0$ giving $x = -12,1$

$x^2+11x+29=-41$ giving $x^2+11x+70=0$ does not have real root

2020/008) A right angled triangle has perimeter 40 m and area 60 $m^2$ Find the lengths of the sides of the triangle.

Let the 3 sides be a,b & c, where c is the hypotenuse and without loss of generality $a>b$.
We have as it is right angled triangle
$a^2+b^2 = c^2\cdots(1)$
As perimeter is 40 we have a+b+c = 40 and hence
$a+b= 40 -c\cdots(2)$
Area is 60 so we have
$\frac{ab}{2} = 60$ or $ab=120\cdots(3)$
From (2) and (3) $(a+b)^2 - 2ab = (40-c)^2 - 240 $
Or $a^2+b^2 = c^2 - 80c + 1600 - 240 = c^2 - 80c + 1360 $
Or $80c-1360 = c^2-(a^2+b^2) = 0$ sing (2)
Or c = 17
Puttng c =17 in (2) we get
$a+b= 40-17 = 23$
So $(a+b)2^ = 23^2$
Using above and (3) we get $(a-b)^2 = (a+b)^2 - 4ab = 23^2 - 4 * 120 = 49$
Or a-b = 7\cdots(4)
From (2) and (4) a = 15 and b = 8
So sides of triangle are 15m,8m,17m


This question I picked from https://in.answers.yahoo.com/question/index?qid=20200310093657AAFSsOP