If
\(\dfrac{1}{1*2} + \dfrac{1}{2*3} + \dfrac{1}{3*4} + \ldots + \dfrac{1}{n*(n + 1)} = \dfrac{99}{100}\),
what is \(n\)?
\(\begin{array}{|rcll|} \hline \dfrac{99}{100} &=& \dfrac{1}{1*2} + \dfrac{1}{2*3} + \dfrac{1}{3*4} + \ldots + \dfrac{1}{n*(n + 1)} \\ &=& \sum \limits_{m=1}^n \dfrac{1}{m*(m + 1)} &| \quad \dfrac{1}{m*(m + 1)} = \dfrac{1}{m}-\dfrac{1}{m + 1} \\ &=& \sum \limits_{m=1}^n \left( \dfrac{1}{m}-\dfrac{1}{m + 1} \right) \\ &=& \sum \limits_{m=1}^n \dfrac{1}{m} -\sum \limits_{m=1}^n \dfrac{1}{m + 1} \\ &=& \sum \limits_{m=1}^n \dfrac{1}{m} -\sum \limits_{m=2}^{n+1} \dfrac{1}{m} \\ &=& \sum \limits_{m=1}^n \dfrac{1}{m} -\sum \limits_{m=2}^{n} \dfrac{1}{m} - \dfrac{1}{n+1} \\ &=& 1 \underbrace{+ \sum \limits_{m=2}^n \dfrac{1}{m} -\sum \limits_{m=2}^{n} \dfrac{1}{m}}_{=0} - \dfrac{1}{n+1} \\ \mathbf{ \dfrac{99}{100} } &=& \mathbf{ 1 - \dfrac{1}{n+1} } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{ \dfrac{99}{100} } &=& \mathbf{ 1 - \dfrac{1}{n+1} } \\\\ \dfrac{1}{n+1}&=&1- \dfrac{99}{100} \\\\ \dfrac{1}{n+1}&=& \dfrac{1}{100} \\\\ n+1 &=& 100 \\ n &=& 100-1 \\ \mathbf{n} &=& \mathbf{99} \\ \hline \end{array} \)
The volume of a cube is nuemrically equal to the surface area of a cube. What is the side length of the cube?
The rectangle in the diagram has an area equal to 640 \text{cm}^2cm2.
Points B and F are midpoints of sides AC and AE, respectively.
What is the area of triangle BDF in \text{cm}^2?cm2?
AM.
Do we not have to divide by 3 ?
Find \((1^2 + 1*2 + 2^2) + (2^2 + 2*3 + 3^2) + \ldots + (99^2 + 99*100 + 100^2)\).
\(\begin{array}{|rcll|} \hline && \mathbf{(1^2 + 1*2 + 2^2) + (2^2 + 2*3 + 3^2) + \ldots + (99^2 + 99*100 + 100^2)} \\ &=& \sum \limits_{n=1}^{99} \left( n^2+n(n+1)+(n+1)^2 \right) \\ &=& \sum \limits_{n=1}^{99} \left( n^2+n^2+n +n^2+2n+1 \right) \\ &=& \sum \limits_{n=1}^{99} \left( 3n^2+3n+1 \right) \\ &=& \sum \limits_{n=1}^{99} \left( 3n^2 \right) +\sum \limits_{n=1}^{99} \left( 3n\right) +\sum \limits_{n=1}^{99} \left( 1 \right) \\ &=& 3\sum \limits_{n=1}^{99} \left( n^2 \right) +3\sum \limits_{n=1}^{99} \left( n\right) + \underbrace{\sum \limits_{n=1}^{99} \left( 1 \right)}_{= 99*1=99} \\ &=& 3\sum \limits_{n=1}^{99} \left( n^2 \right) +3\sum \limits_{n=1}^{99} \left( n\right) + 99 \\\\ && \boxed{\sum \limits_{n=1}^m (n^2) = \dfrac{m(m+1)(2m+1)}{6} \\ \sum \limits_{n=1}^m (n ) = \dfrac{m(m+1)}{2} } \\\\ &=& 3 *\left( \dfrac{99*(99+1)*(2*99+1)}{6}\right) +3 *\left( \dfrac{99*(99+1)}{2}\right) + 99 \\ &=& 3 *\left( \dfrac{99*100*199}{6}\right) +3 *\left( \dfrac{99*100}{2}\right) + 99 \\ &=& 99*50*199 + 3 * 99* 50 + 99 \\ &=& \mathbf{999999} \\ \hline \end{array}\)
In a large equilateral triangle, we draw the incircle. We then draw a smaller equilateral triangle in the incircle.
What is the ratio of the area of the smaller equilateral triangle to that of the larger equilateral triangle?
△ABC is a right triangle and line segment AD is an altitude. If AB = 10 and BD = 8 then what is the area of \△ABC?
Hi Tom,
One solution.
People shrink when they get older because the cartilage between their vertebrae is compacted.
Find the minimum value of \(\dfrac{a}{b} + \dfrac{b}{a}\), where a and b are positive real numbers.
\(\mathbf{\huge{AM \geq GM }}\)
\(\begin{array}{|rcll|} \hline \dfrac{a^2+b^2}{2} &\geq& \sqrt{a^2b^2} = ab \\ \dfrac{a^2+b^2}{2} &\geq& ab \quad | \quad * \dfrac{2}{ab} \\ \dfrac{a^2+b^2}{ab} &\geq& 2 \\ \dfrac{a^2}{ab}+\dfrac{b^2}{ab} &\geq& 2 \\ \mathbf{\dfrac{a}{b}+\dfrac{b}{a}} &\geq& \mathbf{2} \\ \hline \end{array}\)
The minimum value of \(\dfrac{a}{b} + \dfrac{b}{a}\) is 2
If x, y, z are positive real numbers such that xyz = 8, find the minimum value of x + y + z.
\(\mathbf{\huge{AM \geq GM }} \)
\(\begin{array}{|rcll|} \hline \dfrac{x+y+z}{3} &\geq& \sqrt[3]{xyz} = \sqrt[3]{8} \\ \dfrac{x+y+z}{3} &\geq& \sqrt[3]{8} \\ \dfrac{x+y+z}{3} &\geq& \sqrt[3]{2^3} \\ \dfrac{x+y+z}{3} &\geq& 2 \quad | \quad *3 \\ \mathbf{ x+y+z } &\geq& \mathbf{6} \\ \hline \end{array} \)
The minimum value of \(x + y + z\) is 6
Thank you, Guest !
Since the first 14 terms are:
2,5,8,11,14,17,20,23,26,29,32,35,38,41
We take the sum of the first one and the last one:
43, and we multiply that by 7,
And we get 43*7.
Can you figure it out from there?
The probability that you now have two black balls is 2/9 + 4/27 = 10/27.
If we choose the multi-event spectrums, such as rose, rose plus amber, rose plus amber plus emerald, and all possible 2^4 = 16 choices, we'll get a probability of 4/16 - c(4,3)*3/16*15 + c(4,2)*2/16*15*14 - c(4,1)*1/16*15*14*13, which works out to 2/15.
(a) 101^2 - 97^2 + 93^2 - 89^2 + ... + 5^2 - 1^2 = 101 + 97 + 93 + 89 + ... + 5 + 1 = 1326.
(b) (a + (2n + 1)d)^2 - (a + (2n)d)^2 + ... + (a + d)^2 - a^2 = (a + (2n + 1) d + (a + 2nd + ... + a + d) + a = n(3a + (n + 2)d).
How many times do you flip it ??!!!!
Just add them up!
2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41 !!!!!
Using difference of squares, the sum works out to 2(a + (n + 1)d).
8075! = has 2016 trailing zeros.
floor(x/5) + floor(x/5^2) + floor(x/5^3) + floor(x/5^4) =2016, solve for x
x =8075!.
The least positive integer n = 2.
For what range? up to 100? 1000? 1000000?.......etc.
Product = -24 x -23 x -22..........x 0 x 1 x 2 x 3........25 =0
Sum = -24 + -23 + -22..........+ 0 +1 + 2 + 3........ 25 =25
This is how I did it:
All angles add up to 180.
To find the largest angle, we guess and check on our way down from 180.
169
144
121
100
...
The other two angles must add up to the supplement of the numbers above and are perfect squares.
8 x 125 =1,000
336 6679008978, 7833 3456782289, 897 7790114478, Total = 6166680 3779205000
336 6679008978 mod 9999 =0, 7833 3456782289 mod 9999 =0, 897 7790114478 mod 9999 =0