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T1=2, T2=6 and T3=12. Determine the nth term formula.

\(\begin{array}{|rcll|} \hline t_1 = 1 \cdot 2 &=& 2 \\ t_2 = 2 \cdot 3 &=& 6 \\ t_3 = 3 \cdot 4 &=& 12 \\ \ldots \\ t_n &=& n(n+1) \\ \hline 420 &=& n(n+1) \\ 420 &=& n^2+n \\ n^2+n -420 &=& 0 \\ n&=& \dfrac{-1\pm\sqrt{1-4\cdot(-420)}}{2} \\ n&=& \dfrac{-1\pm\sqrt{1681}}{2} \\ n&=& \dfrac{-1\pm 41}{2} \\ n&=& \dfrac{-1 {\color{red}+} 41}{2} \quad | \quad n > 0\ ! \\ \mathbf{n} &\mathbf{=}& \mathbf{ 20 }\\ \hline \end{array}\)

\(\mathbf{t_{20} = 20\cdot 21 = 420}\)

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.

Let Fibonacci number sequence is:

\(\begin{array}{|rcll|} \hline \ldots \\ f_{-1} &=& 1 \\ f_{0} &=& 0 \\ f_{1} &=& 1 \\ f_{2} &=& 1 \\ f_{3} &=& 2 \\ f_{4} &=& 3 \\ f_{5} &=& 5 \\ f_{6} &=& 8 \\ f_{7} &=& 13 \\ f_{8} &=& 21 \\ f_{9} &=& 34 \\ f_{10} &=& 55 \\ f_{11} &=& 89 \\ f_{12} &=& 144 \\ \ldots \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_n &=& f_{n-2}a_1+f_{n-1}a_2 \\ s_n &=& f_na_1+ \left(f_{n+1}-1 \right)a_2\\ \hline a_{7} = f_5a_1+f_6a_2 &=& 6 \\ f_6a_2 &=& 6-f_5a_1 \\ a_2 &=& \frac{6-f_5a_1}{f_6} \\\\ s_{10} &=& f_{10}a_1+ \left(f_{11}-1\right)a_2\\ &=& f_{10}a_1+ \left(f_{11}-1\right)\left(\frac{6-f_5a_1}{f_6}\right)\\ &=& f_{10}a_1+ \dfrac{6\left(f_{11}-1\right)}{f_6}- \dfrac{f_5\left(f_{11}-1\right)}{f_6}a_1 \\ &=& \left(f_{10}- \dfrac{f_5\left(f_{11}-1\right)}{f_6}\right)a_1+ \dfrac{6\left(f_{11}-1\right)}{f_6} \\ &=& \left(55- \dfrac{5\left(89-1\right)}{8}\right)a_1+ \dfrac{6\left(89-1\right)}{8} \\ &=& \left(55- \dfrac{5\cdot 88}{8}\right)a_1+ \dfrac{6\cdot 88}{8} \\ &=& (55- 55 )a_1+ 6\cdot 11 \\ &=& 0+66 \\ \mathbf{s_{10}} &\mathbf{=}& \mathbf{ 66 } \\ \hline \end{array}\)

Note that

3^2 - 1 =  8 = 2*4

5^2 - 1 = 24 = 4*6

7^2 - 1 = 48 =  6*8

So  we have  for n = 1 to infinity

6 *    1 / [ 2n * (2n + 2) ]  =  6 *   1 / [ 4 (n)(n+ 1) ]   =  (3/2) * 1/ [ n * (n + 1) ]

Note that     1  / [ n * (n + 1) ]   can be written as    1 /  n - 1/{n + 1]

So we have

(3/2)  [ ( 1 - 1/2 ) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + ...... ]    

All the terms cancel  except  for the first term , 1,   and the last term , -1/(n + 1)

And when n is large  - 1 (n + 1) → 0

So....the sum is

(3/2) (1)  =

3/2

help

\(\large{\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots}\)

\(\begin{array}{|rcll|} \hline && \dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots \\\\ &=& \dfrac{6}{(3-1)(3+1)}+\dfrac{6}{(5-1)(5+1)}+\dfrac{6}{(7-1)(7+1)}+\dfrac{6}{(9-1)(9+1)}+\cdots \\\\ &=& \dfrac{6}{2\cdot 4}+\dfrac{6}{4\cdot 6}+\dfrac{6}{6\cdot 8}+\dfrac{6}{8\cdot 10}+\cdots \\\\ &=& 6\left( \dfrac{1}{2\cdot 4}+\dfrac{1}{4\cdot 6}+\dfrac{1}{6\cdot 8}+\dfrac{1}{8\cdot 10}+\cdots \right) \\\\ &=& 6\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{2k(2k+2)} \\\\ &=& 6\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{4k(k+1)} \\\\ &=& \dfrac{6}{4}\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{k(k+1)} \\\\ &=& \dfrac{3}{2}\cdot \sum \limits_{k=1}^{\infty} \dfrac{1}{k(k+1)} \quad | \quad \boxed{\dfrac{1}{k(k+1)} = \dfrac{1}{k} - \dfrac{1}{k+1} } \\\\ &=& \dfrac{3}{2}\cdot \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k} - \dfrac{1}{k+1} \right) \\\\ &=& \dfrac{3}{2}\cdot \left[\sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k+1} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot \left[1+\sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=1}^{\infty} \left(\dfrac{1}{k+1} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot \left[1+\sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) - \sum \limits_{k=2}^{\infty} \left(\dfrac{1}{k} \right) \right] \\\\ &=& \dfrac{3}{2}\cdot (1+0) \\\\ &=& \dfrac{3}{2} \\ \hline \end{array}\)

\(\text{If you grind through it using }a_1,~a_2 \text{ as the first two sequence elements you get}\\ s =\left\{a_1,a_2,a_1+a_2,a_1+2 a_2,2 a_1+3 a_2,3 a_1+5 a_2,5 a_1+8 a_2,8 a_1+13 a_2,13 a_1+21 a_2,21 a_1+34 a_2\right\}\\ \text{Summing we get }\\ \sum \limits_{n=1}^{10}~s_k = 55 a_1+88 a_2\\ s_7=5 a_1+8 a_2 = 6\\ a_2 = \dfrac{1}{8} \left(6-5 a_1\right)\\ \text{dumping this into the sum we find (like magic!)}\\ \sum \limits_{n=1}^{10}~s_k = 66\)

Notice that

1   + 2 - 3 = 0

4 + 5 -   6 = 3

7 + 8 - 9 =   6

So....the first 24 terms will be groups of 3 and there will be 8 of these groups 

And the last group will sum to  3(8-1) = 21

So the sum of these 8 groups will be    [ 0 + 21] * (8/2)  =  21 * 4 =  84

And the 25th term = 25

So...the sum is

84 + 25 =

109

sumfor(n, 1, infinity, 6/((2*n + 1)^2 - 1)) = 3 / 2

OK...we got something we can work with...

If AFM is 60°.....then angle RFM =  30°

And the top left triangle is a 30-60-90 right triangle

And the side opposite the 90°  angle is the side of 7

And the side opposite the 30° angle is 1/2 the side length = 1/2*7 =  3.5

So... 1/2 of diagonal MA =  1/2(7) = 3.5

So....the whole length of AM is twice this = 7 units

And angle FMR is supplemental to angle AM =  120°

Then angle AMR is 1/2 of this = 60°

And the bottom left triangle is also 30-60-90

The side opposite  the 60° angle  = 3.5sqrt(3) units

And this is 1/2 the length  of diagonal  FR

So..its length = 2 *3.5sqrt(3)  =  7sqrt (3) units

Fixed, thanks Melody.

First, set x=3.444

Then, 10x=34.4444

9x=31 and x=31/9=3 4/9.

3  4/9  =   31 / 9

Don't think we can solve this one without some other info, Josey......

See here :  https://web2.0calc.com/questions/algebra-help_37

sin T =   opp/ hyp = n / c

cos R =  adj / hyp =  n / c

So...equal!!!

Goodnight!!

1. The max angle is given by this :

arctan (1/12) = angle  ≈  4.76°

2. The tan is the rise over the run.....hence 1 ft rise / 12 foot run

The tan inverse (arctan) will tell us what an angle with this rise and run is 

3. Ramp length can be found by this

sin (4.76°) = 12 / ramp length        .....rearrange as

ramp length =    12/ sin (4.76°) ≈   144.6 ft

4. The sine gives the ratio of the opposite side from the angle to the hypotenuse which is the ramp length

the answer is B

Simplifying

         3(4x + -1) + -7x = 17

Reorder the terms:

        3(-1 + 4x) + -7x = 17 (-1 * 3 + 4x * 3) + -7x = 17 (-3 + 12x) + -7x = 17

Combine like terms:

       12x + -7x = 5x -3 + 5x = 17

Solving

       -3 + 5x = 17

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right. Add '3' to each side of the equation.

       -3 + 3 + 5x = 17 + 3

Combine like terms:

       -3 + 3 = 0 0 + 5x = 17 + 3 5x = 17 + 3

Combine like terms:

       17 + 3 = 20 5x = 20

Divide each side by '5'. x = 4 Simplifying x = 4

:D

hellllllo :D

Maximum slope for hand-propelled wheelchair ramps should be 1" of rising to every 12" of length (4.8-degree angle; 8.3% grade).

Maximum slope for power chairs should be 1.5" rise to 12" length (7.1-degree angle; 12.5% grade).

Minimum width should be 36" (inside rails) - (48" is ideal).

100 Hz x .30 m = 30 m/s

Thank you for making me feel a lot better about this problem! btw this is for my algebra 2 class and we tried using a graphing calculator and this didn't work so I ended up here trying to find better sources of information Thanks again! 

Ah, yes...missed that one!!

Good job!!!