A skier has decided that on each trip down a slope, she will do 2 more jumps than before. On her first trip she did 6 jumps. Derive the sigma notation that shows how many total jumps she attempts from her fourth trip down the hill through her twelfth trip. Then solve for how many total jumps she attempts from her fourth trip down the hill through her twelfth trip.

I tried to answer, but I don't really know how to even set it up... :/

Respuesta :

I'll just manually compute this one. Every attempt she will add 2 more jumps.

Attempt           Jumps       Accumulated Jumps
1                          6                         6
2                          8                       14
3                         10                      24
4                         12                      36
5                         14                      50
6                         16                      66 
7                         18                      84
8                         20                    104
9                         22                    126
10                       24                    150
11                       26                    176
12                       28                    204

Total jumps from 4th trip through 12th trip: 204 - 24 = 180 jumps

The correct answers are:

[tex] \Sigma_{n=4}^{12} 6+2(n-1)
\\
\\=180 [/tex]

Explanation:

Since she is adding two more jumps every time she goes down hill, this is an arithmetic sequence. The general form of an arithmetic sequence is

[tex] a_n=a_1+d(n-1),
\\
\text{where } a_n \text{represents the nth term, } a_1 \text{represents the first term, and n represents the term number} [/tex]

Since we want the number of jumps on her 4th through 12th trips, we will set n in the summation from 4 to 12. n=4 goes at the bottom of Σ and 12 goes at the top, to represent the values we are interested in.

Beside this, we write our general form. The first term is 6 and d, the common difference, is 2. This gives us 6+2(n-1) beside the summation:

[tex] \Sigma_{n=4}^{12} 6+2(n-1) [/tex]

To evaluate this, we substitute the values 4, 5, 6, 7, 8, 9, 10, 11 and 12 in for n, adding all of the values together:

6+2(4-1)+6+2(5-1)+6+2(6-1)+6+2(7-1)+6+2(8-1)+6+2(9-1)+6+2(10-1)+6+2(11-1)+6+2(12-1)

=6+6+6+8+6+10+6+12+6+14+6+16+6+18+6+20+6+22

=180