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I don't understand how that solves the question, can you please elaborate?

thanks!

Hi good people!,
How do I go about finding the general term for the following sequence:
{1\over3};{8\over9};1;{64\over81}

\(\displaystyle {1\over3};{8\over9};1;{64\over81}\)

\(\huge{ \begin{array}{|rcll|} \hline a_1 &=& {1\over3} \\\\ a_2 &=& {8\over9} \\\\ a_3 &=& 1 \\\\ a_4 &=& {64\over81} \\\\ \ldots \\\\ a_n &=& \dfrac{n^3}{3^n} \\ \hline \end{array} }\)

Try the following:

\(1^3\frac{1}{3}; 2^3\frac{1}{3}^2;3^3\frac{1}{3}^3;4^3\frac{1}{3}^4;...etc\)

3² + 4² = 5²

Let n be any other integer, multiply the above by n²

(3n)² + (4n)² = (5n)²

There are an infinite number of integer solutions: x = 3n, y = 4n and N = (5n)² is also an integer.

The product of $3t^2+5t+a$ and $4t^2+bt-2$ is $12t^4+26t^3-8t^2-16t+6$. What is $a+b$?

\(\begin{array}{|rcll|} \hline && (3t^2+5t+a)(4t^2+bt-2) \\ &=& 12t^4+3bt^3-6t^2+20t^3+5bt^2-10t+4at^2+abt-2a \\ &=& 12t^4+(3b+20)t^3 +(-6+5b+4a)t^2 +(-10+ab)t -2a \\ \hline \text{compare}&=&12t^4+26t^3-8t^2-16t+6 \\\\ -2a &=& 6 \quad | \quad : (-2) \\ \mathbf{a} & \mathbf{=} & \mathbf{-3} \\\\ 3b+20 &=& 26 \\ 3b &=& 6 \\ \mathbf{b} & \mathbf{=} & \mathbf{2} \\\\ a+b &=& -3+2 \\ \mathbf{a+b} & \mathbf{=} & \mathbf{-1} \\ \hline \end{array} \)

Let  S = 1^2/(1.3) + 2^2/(3.5) + 3^2/(5.7) + 4^2/(7.9) +............+ 500^2/(999.1001).

What is the sum of S?

\(\begin{array}{|rcll|} \hline S &=& \dfrac{1^2}{1\cdot 3} + \dfrac{2^2}{3\cdot 5} + \dfrac{3^2}{5\cdot 7} + \dfrac{4^2}{7\cdot 9} + \ldots + \dfrac{500^2}{999\cdot 1001} \\\\ S &=& \sum \limits_{n=1}^{500} \dfrac{n^2}{(2n-1)(2n+1)} \\\\ && \boxed{\dfrac{1}{(2n-1)(2n+1)}=\dfrac12\cdot \left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right)} \\\\ S &=& \sum \limits_{n=1}^{500} n^2\dfrac12 \left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right) \\\\ S &=& \dfrac12 \sum \limits_{n=1}^{500} n^2 \left( \dfrac{1}{2n-1} - \dfrac{1}{2n+1} \right) \\\\ S &=& \dfrac12 \sum \limits_{n=1}^{500} \left( \dfrac{n^2}{2n-1} - \dfrac{n^2}{2n+1} \right) \\\\ 2S &=& \sum \limits_{n=1}^{500} \left( \dfrac{n^2}{2n-1}\right) - \sum \limits_{n=1}^{500} \left( \dfrac{n^2}{2n+1} \right) \\\\ 2S &=& \dfrac{1^2}{1\cdot2-1} + \sum \limits_{n=2}^{500} \left(\dfrac{n^2}{2n-1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{2\cdot 500 + 1} \\\\ 2S &=& 1 + \sum \limits_{n=2}^{500} \left(\dfrac{n^2}{2n-1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{(n+1)^2}{2(n+1)-1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{(n+1)^2}{2n+1}\right) - \sum \limits_{n=1}^{499} \left(\dfrac{n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{(n+1)^2-n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{n^2+2n+1-n^2}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(\dfrac{2n+1}{2n+1}\right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + \sum \limits_{n=1}^{499} \left(1 \right) -\dfrac{500^2}{1001} \\\\ 2S &=& 1 + 499 -\dfrac{500^2}{1001} \\\\ 2S &=& 500-\dfrac{500^2}{1001} \\\\ 2S &=& 500 \cdot \left(1-\dfrac{500}{1001} \right) \\\\ S &=& 250 \cdot \left(\dfrac{1001-500}{1001} \right) \\\\ S &=& \dfrac{250\cdot 501}{1001} \\\\ \mathbf{S} & \mathbf{=} & \mathbf{\dfrac{125250}{1001}} \\ \hline \end{array}\)

Find the value of $y$ such that $9^y = \dfrac{3^5\cdot 9^6}{27^3}$.

\(\begin{array}{|rcll|} \hline 9^y &=& \dfrac{3^5\cdot 9^6}{27^3} \\\\ 9^y &=& \dfrac{3^5\cdot 9^6}{(3\cdot 9)^3} \\\\ 9^y &=& \dfrac{3^5\cdot 9^6}{3^39^3} \\\\ 9^y &=& 3^{5-3}\cdot 9^{6-3} \\ 9^y &=& 3^{2}\cdot 9^{3} \\ 9^y &=& 9\cdot 9^{3} \\ 9^y &=& 9^1\cdot 9^{3} \\ 9^y &=& 9^{3+1} \\ 9^{{\color{red}y}} &=& 9^{{\color{red}4}} \\\\ \mathbf{y} & \mathbf{=} & \mathbf{4} \\ \hline \end{array}\)

Each face of a cube is assigned a different integer.
Then each vertex is assigned the sum of the integer values on the faces that meet at the vertex.
Finally, the vertex numbers are added.
What is the largest number that must divide the final sum for every possible numbering of the faces?

Each face of a cube is assigned a different integer.
Then each vertex is assigned the sum of the integer values on the faces that meet at the vertex.
Finally, the vertex numbers are added.
What is the largest number that must divide the final sum for every possible numbering of the faces?

\(\begin{array}{|c|c|} \hline \text{face} & \text{assigned with integer} \\ \hline 1 & a \\ 2 & b \\ 3 & c \\ 4 & d \\ 5 & e \\ 6 & f \\ \hline \end{array}\)

\(\begin{array}{|r|r|l|} \hline \text{Vertex} & \text{faces} & \text{sum}\\ \hline 1 & a,d,e & a+d+e \\ 2 & a,b,e & a+b+e \\ 3 & b,c,e & b+c+e \\ 4 & c,d,e & c+d+e \\ 5 & a,d,f & a+d+f \\ 6 & a,b,f & a+b+f \\ 7 & b,c,f & b+c+f \\ 8 & c,d,f & c+d+f \\ \hline &\text{sum } =&(a+d+e)+(a+b+e)+(b+c+e)+(c+d+e)\\ & &+(a+d+f)+(a+b+f)+(b+c+f)+(c+d+f) \\ & = & 4a+4b+4c+4d+4e+4f \\ & = & {\color{red}4}(a+b+c+d+e+f) \\ \hline \end{array}\)

The largest number that must divide the final sum for every possible numbering of the faces is 4

Let f(y) = y^4 -3y^3 +y - 3 and g(y) = y^3 + 7y^2 -2. Find $f(y) + g(y)$.

Write your answer as a polynomial with terms of decreasing degree.

\(\begin{array}{|rcrcrcrcrcr|} \hline f(y) &=& y^4 &-& 3y^3 & & &+& y &-& 3 \\ g(y) &=& & & y^3 &+& 7y^2 & & &-& 2 \\ \hline f(y)+g(y) &=& y^4 &-& 2y^3 &+& 7y^2 &+& y &-& 5 \\ \hline \end{array}\)

Let S = 1^2/(1.3) + 2^2/(3.5) + 3^2/(5.7) + 4^2/(7.9) +.....+ 500^2/(999.1001)
Each term of S is of the form:n^2 /[(2n - 1)(2n + 1)]=1/4[n/(2n-1)+ n/(2n+1)]. Thus:
S = 1/4*∑[n/(2n-1)+ n/(2n+1)], n= 1 to 500
4S =∑[n/(2n-1)+ n/(2n+1)], n= 1 to 500=1/1+1/3+2/3+2/5+3/5+3/7+......+500/999+500/999. And since:
[n/(2n-1)+ (n+1)/(2n+1)] = 1, we can combine all such terms to get: 4S = 1 + 499 + 500/1001 =501,000/1,001, and:
S =125,250/1,011 =125 +125/1,011

 9^y = (3^5×9^6)/27^3 solve for y

9^y = 6,561             factor both sides

3^2y = 3^8             equate the exponents

2y = 8                       divide both sides by 2

y = 8/2

y = 4

Huh? Can someone else help too, like a moderator as well?

x^2 + y^2 =N. This is the equation of circle!!

PLEASE post  readable questions.

I got the answer, no worries

Yep...you are correct...I forgot that we  need to  find the integer values that make the inequality untrue...-4 should be included!!!

We can solve this by inspection

Note that when  x = 1   the expression  under the entire radical  is 0

And x cannot be smaller than this   (note if it were, for example, 0,   1 - √2  under the radical would be  negative

And  x  cannot be  larger than 4   because    [2 - √x ]   would result in a negative quantity under a radical

So...the domain  is  [ 1, 4 ]

Here's a graph : https://www.desmos.com/calculator/xxzra4wyai

Lol! Maybe i should! BTW, -4 is included! Thanks!

Would this set satisfy the conditions given?

1-      6 + 3 + 2 + 1 = 12
2-     5 + 4 + 2 + 1 = 12
3-     5 + 3 + 3 + 1 = 12
4-     5 + 3 + 2 + 2 = 12
5-     5 + 3 + 2 + 1 + 1 = 12
6-     4 + 4 + 3 + 1 = 12
7-     4 + 4 + 2 + 2 = 12
8-     4 + 4 + 2 + 1 + 1 = 12
9-     4 + 3 + 3 + 2 = 12
10-   4 + 3 + 3 + 1 + 1 = 12
11-    4 + 3 + 2 + 2 + 1 = 12
12-   4 + 3 + 2 + 1 + 1 + 1 = 12

No prob...maybe they should call Me "Lightning,"  huh ??

5x^2 + 19x + 16  > 20      subtract 20 from both sides

5x^2 + 19x - 4   > 0       let's set this to 0

5x^2 + 19x - 4  = 0     factor, if we can

(5x -1 ) ( x + 4)  = 0      set each factor to 0 and solve for x and we  get

x  =1/5     and  x = -4

Note that  we have three possible intervals that make the original inequality  true

(-inf, -4)  or  ( - 4, 1/5)  or  ( 1/5, inf )

Normally, in a quadratic...if the middle interval  makes the inequality untrue, the  outside intervals will make it  true

Testing a point in the middle interval  - I'll pick  x  = 0 -  and subbing it into the original inequality  produces  16 > 20

So....the  middle interval makes the inequality untrue....and the integers in this interval  are  -3, -2, -1 and 0  [ 4 values]

(Note that we  can't include -4 since it is not part of the interval itself)

See the graph, here : https://www.desmos.com/calculator/dkmaryyfcp

Thank you Chris for getting back to me so fast!

We have no restriction on the domain  for the second function  because we can take the cube root of any real number

However....for the first function,  x cannot be  < 2  because we would have a negative square root which does not produce a real  value

So......we must take the most restrictive domain  for both functions....this is  [ 2,  infinity )

Here's a graph : https://www.desmos.com/calculator/ant19juary

Given that the point (4,7) is on the graph of y=f(x), there is one point that must be on the graph of 2y=3f(4x)+5. What is the sum of coordinates of that point?

2y  = 3f(4x) + 5       divide through by  2

y = (3/2) f(4x) + 5/2

The  "4x"  part  compresses the graph horizontally by a factor of 4

The "3/2"  part vertically  stretches the graph by a factor  of 3/2

The "5/2"  part  shifts the resulting point up by  5/2 units

Applting the first two  to (4,7)  we get  (4*(1/4) , 7* (3/2) ) =  ( 1, 21/2)

Applying the last one, we arrive at   ( 1, 21/2 + 5/2)  = (1, 26/2)  = ( 1,13)

The sum of the coordinates  is  1 + 13   = 14