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Thank you for the hint. I am still a bit confused. I am guessing the answer is 40-22=18. Not sure about figuring out the 22 part. Is it a trick question or am I just not "getting it"?

 2)In a sequence of ten terms, each term (starting with the third term) is equal to the sum of the two previous terms. The seventh term is equal to 6. Find the sum of all ten terms.

a1 = a1

a2 = a2

a3 = a1 + a2

a4 = a2 + a3    =  a1 + 2a2 

a5 = a3 + a4    =   2a1 + 3a2

a6 =  a4 + a5    = 3a1 + 5a2

a7 =  a5 + a6    =  5a1 + 8a2  = 6    →   a1  =  (1/5) (6 - 8a2)   (1)

a8 =  8a1 + 13a2

a9 = 13a1 + 21a2

10 = 21a1 + 34a2

So....the sum of the first ten terms =   55a + 88a2

Subbing (1) into this.....the sum is

55(1/5) (6 - 8a2) + 88a2  =

11(6 - 8a2) + 88a2  =

66 - 88a2 + 88a2 =

66

My calculator kept giving me -0.004784853  but it's probably just because I was putting it in wrong.  Thank you.

Oops. Thank you for the correction. 

There are 99 sequences to sum up.

The n-th sequence looks like this:

n(n+1) + (n+1)(n+2) + ... + 99(100).

We can solve this by summation:

$\quad n(n+1 ) + (n+1)(n+2) + \cdots + 99(100)\\ =(1\cdot2 + 2\cdot 3 + ... + 99(100)) - (1\cdot 2 + 2\cdot 3 + ... + n(n-1))\\ =\displaystyle \sum^{99}_{k=1} (k^2 + k) - \sum^{n-1}_{k=1}(k^2 + k)\\ = \dfrac{99(100)(199)}{6} + \dfrac{99(100)}{2} - \dfrac{n(n-1)(2n-1)}{6} - \dfrac{n(n-1)}{2}\\ = -\dfrac{n^3}{3} + \dfrac{n}{3} + 333300$

Now we need to find the sum of this expression from n = 1 to p, where p is some positive integer.

$\displaystyle\sum^{p}_{n=1} \left(\dfrac{-n^3}{3} + \dfrac{n}{3} + 333300\right)\\ = \dfrac{p(-p^3-2p^2+p+3999602)}{12}$

We sum this from p = 1 to p = 100 and we get 1,508,082,510. 

:)

To Guest: The first sequence is (2 + 3 + 4 + ... + 100) + (6 + 8 + 10 + ... + 200) + ... + (98 * 99 + 98*100 + 99*100).

$\tan Y = \dfrac{\text{XZ}}{\text{XY}} = \dfrac{\sqrt{25^2 - 24^2}}{24} = \dfrac{7}{24}$

$x^5y^2 - x^3 y + 5xy^3 = 0\\ 5x^4 y^2 \dfrac{dx}{dy} + 2x^5y-3x^2y\dfrac{dx}{dy} -x^3 +5y^3 \dfrac{dx}{dy} + 15xy^2 = 0\\ (5x^4y^2 -3x^2y+5y^3)\cdot\dfrac{dx}{dy} = x^3-2x^5y-15xy^2 \\ \dfrac{dx}{dy} = \dfrac{x(x^2 - 2x^4y-15y^2)}{y(5x^4y-3x^2+5y^2)}$

Max: The last one forms an Arithmetic Sequence as follows:

199, 195, 191, 187, 183......and so on for 50 terms

Sum =50/2*[2*199 + (50 -1) * - 4]

Sum =5050

Maximum occurs at $x_1 = x_2 = x_3 = \cdots = x_{101} = \dfrac{1}{\sqrt{101}}$.

Maximum value is $100\left(\dfrac{1}{\sqrt{101}}\right)^2 =\dfrac{100}{101}$.

Notes: The equation $x_1^2 + x_2^2 + \cdots + x_{101}^2 = 1$ represents a 101-sphere with radius 1.

:)

$x = \{1,11\} \implies (x-1)(x-11) = 0\implies x^2 - 12 x + 11 = 0\\ x = \{2,10\} \implies (x-2)(x-10) = 0 \implies x^2 - 12 x + 20 = 0\\ x = \{3,9\} \implies (x-3)(x-9) = 0\implies x^2 - 12x + 27 = 0\\ x = \{4,8\} \implies (x-4)(x-8) = 0\implies x^2 - 12x + 32 = 0\\ x = \{5,7\} \implies (x-5)(x-7) = 0\implies x^2 -12x + 35 = 0\\$

For any adjacent terms (2n)² and (2n - 1)², difference = 4n - 1.

$100^2 - 99^2 + 98^2 - 97^2+\cdots + 2^2 - 1^2\\ = 199 + 195 + 191+\cdots + 3\\ =\dfrac{(3+199)(\frac{199-3}{4}+1)}{2}\\ =5050$

2.

Assume $a_1 = x,\quad a_2 = y$.

$a_3 = x + y\\ a_4 = x+2y\\ a_5 = 2x+3y\\ a_6 = 3x + 5y\\ a_7 = 5x + 8y\\ a_8 = 8x + 13y\\ a_9 = 13x + 21y\\ a_{10} =21x+34y$

WLOG, assume y = 0 $\implies x = \dfrac{6}{5}$.

Sum of 10 terms = $\dfrac{6}{5} \left(1+0+1+1+2+3+5+8+13+21\right) = \dfrac{6}{5} \cdot(55) = 66$

1.

$a_1 = 13\\ a_2 = 28\\ a_3 = 520\\ a_4 = 133\\ a_5 = 55\\ a_6 = 250\\ a_7 = 133\\ a_8 = 55\\ a_9 = 250\\ \vdots$

As you see, for all $n \geq 4$, $a_n = \begin{cases}133&&n\equiv1\pmod3\\ 55&& n\equiv 2 \pmod 3\\ 250&& n\equiv 0 \pmod 3\end{cases}$.

$\therefore a_{100} = 133$

The base case n = 0 is obviously correct, as 0⁵ - 0 = 0 is divisible by 30.

Next, we want to prove when n⁵ - n is divisible by 30, (n+1)⁵ - (n+1) is divisible by 30.

To do so, we need to prove that ((n+1)⁵ - (n+1)) - (n⁵ - n) is divisible by 30.

Making use of the fact that a⁵ - b⁵ = (a - b)(a⁴ + a³b + a²b² + ab³ + b⁴), 

((n+1)⁵ - (n+1)) - (n⁵ - n) = ((n+1)⁴ + n(n+1)³ + n²(n+1)² + n³(n+1) + n⁴) - 1 = 5n(n+1)(n²+n+1).

If we can prove that n(n+1)(n²+n+1) is divisible by 6, we are done.

First case: n = 3k for some integers k.

n(n+1)(n²+n+1) = 3k(3k+1)(9k² + 3k + 1).

As 3k is divisible by 3 and (3k+1) is divisible by 2, we proved that n⁵ - n is divisible by 30 when $n\equiv 0 \pmod 3$

Second case: n = 3k + 1 for some integers k.

n(n+1)(n²+n+1) = (3k+1)(3k+2)(9k² + 9k + 3) = 3(3k+1)(3k+2)(3k² + 3k + 1).

As (3k+1) is divislbe by 2, we proved that n⁵ - n is divisible by 30 when $n\equiv 1\pmod 3$.

Last case: n = 3k - 1 for some integers k.

n(n+1)(n²+n+1) = (3k)(3k-1)(9k² - 3k + 1)

When k is even, 3k is divisible by 6.

When k is odd, 3(3k-1) is divisible by 6.

Therefore, n⁵ - n is divisible by 30 when $n\equiv 2\pmod 3$

Therefore, for all n, n⁵ - n is divisible by 30.

Your LaTeX is hard to understand please click on the LaTeX button next to the image one when you compose a document and paste your question in there and select ok

You read the line that says "A052918              a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1)."

That's the index of the sequence and the recursive formula for it.

Then the list below are the members of the sequence.

4.

$\displaystyle \sum^{n}_{k=1} a_k = n(n+1)(n+2)$.

$a_{10} = \displaystyle \sum^{10}_{k=1} a_k - \sum^9_{k=1} a_k = 10(11)(12) - 9(10)(11) = 330$

Thanks for the two answers but for the link on the first one I am confused on how to read it.

3.

$a_1 = 1\\ a_2 = \dfrac{a_1}{1+a_1} = \dfrac{1}{2}\\ a_3 = \dfrac{a_2}{1+a_2} = \dfrac{1}{3}\\ a_4 = \dfrac{a_3}{1+a_3} = \dfrac{1}{4}\\ \vdots$

Obviously, $a_n = \dfrac{1}{n}\quad \forall n \geq 1$.

$\therefore a_{10} = \dfrac{1}{10}$

2.

$\dfrac{6}{3^2-1} + \dfrac{6}{5^2-1} +...\\ = \displaystyle 6\sum^{\infty}_{n=0} \dfrac{1}{(2n+3)^2 - 1}\\ = \displaystyle \dfrac{3}{2}\sum^{\infty}_{n=0} \dfrac{1}{(n+1)(n+2)}\\ = \displaystyle \dfrac{3}{2}\sum^{\infty}_{n=0} \left(\dfrac{1}{n+1} - \dfrac{1}{n+2}\right)\\ = \displaystyle \dfrac{3}{2}\left(\sum^{\infty}_{n=0} \dfrac{1}{n+1} - \sum^{\infty}_{n=0}\dfrac{1}{n+2}\right)\\ = \displaystyle \dfrac{3}{2}\left(\sum^{\infty}_{n=0} \dfrac{1}{n+1} - \sum^{\infty}_{n=1}\dfrac{1}{n+1}\right)\\ = \dfrac{3}{2}\left(\dfrac{1}{0+1}\right)\\ =\dfrac{3}{2}$

1.

(Wow, this is some really advanced homework)

The generated sequence is 1,5,26,135,701,..., which is OEIS A052918 (Took me a while to find it). And a₁₀ is 2,646,275.

Notes: The general relation can be rewritten as a_n+1 = 5a_n + a_n-1.

https://oeis.org/A052918 

^ Here's a link to the sequence.

a) I'm assuming that you are rolling 2 6-sided dice:

Sample space = 6 x 6 = 36

b) Probability of rolling a sum of 4 is: 3+1, 1+3 and 2+2 =3/36 =1/12

    Probability of rolling a sum > 9 is as follows: 6+4=10, 4+6=10 and 5+5=10, or: 3 / 36 =1/12

    Probability of rolling a sum of 11 =6+5 and 5+6 =11, or: 2/36 = 1/18

    Probability of rolling a sum of 12 = 6+6 = 12, or: 1 /36

So, the overall probability of rolling a sum > 9 is: 1/12 + 1/18 + 1/36 = 1/6

1)  if you replace  x  with   x-3   then

The whole parabola will be translated 3 units over in the positive x direction.

2)   Sketch  the graph.   If you know what an axis of symmetry is then the answer will be easily found.

3)  This about what I said for question 1.

Please no one answer over the top of me. 

Well giving wrong answers isn't the best thing to do because that is a form of lying and were not liers on here its more of we will huide you through it but not give all the time a direct answer...

There were 20% more oranges than grapes. 

$o = (1+0.2)g$ 

When 126 fruits were added, the oranges were increased by 10% and the grapes increased by 9%

$(1+0.1)o + (1+0.09)g = o + g + 126$ 

Plug the first equation into the second and simplify and solve for g

Then use that value to calculate o

Finally find o + g