Series, Sigma, Binomials Jeopardy Template

Arithmetic Series

Geometric Series

Sums of Series

Sigma Notation

Binomial Expansions

100

Is this arithmetic, geometric or neither? If arithmetic, what is the difference? If geometric, what is the ratio?

1/2 + 5/6 + 7/6 + 3/2 + 11/6...

Arithmetic with d=1/3

100

Is this arithmetic, geometric or neither? If arithmetic, what is the difference? If geometric, what is the ratio?

1/3 + 2/15 + 4/75 + 8/375 + ...

Geometric, r = 2/5

100

Find the sum:

-12 + -7 + -2 + ... + 183

t_n = -12 + (n-1)(5) S₄₀ = 40(-12 + 183)/2

183 = -12 + 5n -5 = 20(171)

n=40 = 3,420

100

Expand 10

∑ 3n/(n+1)

n=5

3(5)/(5+1) + 3(6)/(6+1) + 3(7)/(7+1) + 3(8)/(8+1) + 3(9)/(9+1) + 3(10)/(10+1)

= 15/6 +18/7 +21/8 + 24/9 + 36/10 + 30/11

100

What is the 10th row of Pascal's Triangle?

1 10 45 120 210 252 210 120 45 10 1

200

Find the 20th term in the series

1 + 16 + 31 + 46 + ...

t_n= 1 + (n-1)15

=15n-14

t₂₀= 15(20)-14 = 286

200

Find the 8th term in the series: 2 + 10 + 50 + 250 + ....

t_n = 2(5)^n-1

^t₈ = 2(5)⁷ =156,250

200

Find the sum of the first 100 terms of the series:

4 + 7 + 10 + 13 + ...

t₁₀₀ = 4 + (100-1)(3)

= 301

S₁₀₀ = 100(4 + 301)/2

= 50(305) = 15,250

200

Write in sigma notation:

1 + 3 + 5 + ... + 199

100 t_n = 1 + (n-1)2

∑ 2n-1 = 1 + 2n - 2 = 2n - 1

n=1 2n - 1 = 199

2n = 200

n = 100

200

Expand (a + b)⁷

a⁷ + 7a⁶b + 21a⁵b² + 35a⁴b³ + 35a³b⁴ + 21a²b⁵ + 7ab⁶+ b⁷

300

Find the formula for the nth term and then find the 101st term:-2, -11, -20, ...

t_n = -9n + 7

t₁₀₁ =-902

300

Find the formula for the nth term and then find the 10th term: 8, 12, 18, 27, ...

t_n = 8 *(3/2)ⁿ^-1

t₁₀ = 8 * (3/2)⁹ = 307.55

300

Find the sum of the first 9 terms of

t_n = -960(-3/2)^n-1

S₉ = -960(1 - (-3/2)⁹) / 1 - (-3/2)

= -960(1 + 3/2⁹) / 5/2

= -384(1 + 3/2⁹) = -15,146.25

300

Write in sigma notation: 6 -12 + 24 - ... -192

6 t_n= 6(-2)^n-1

∑ 6(-2)^n-1 -192 = 6(-2)^n-1

n=1-32 = (-2)^n-1

(-2)⁵ = (-2)^n-1

6 = n

300

Expand (3x - y)⁵

(3x)⁵ + 5(3x)⁴(-y) + 10(3x)³(-y)² + 10(3x)²(-y)³ + 5(3x)(-y)⁴ + (-y)⁵

= 243x⁵ - 405x⁴y + 270x³y² - 90x²y³ + 15xy⁴ - y⁵

400

Find the position of the last term:

25 + 33 + 41 + ... + 145

t_n=25 + (n-1)8 = 25 + 8n -8 = 17 + 8n

17 + 8n = 145

8n = 128

n=16

400

Find the position of the last term in the series: 3 + 6 + 12 + 24 + ... + 384

t_n = 3(2)^n-1

384 = 3(2)^n-1

128 = 2^n-1

2⁷ = 2^n-1

n=8

400

Find the sum of the series: 1,500 + 1,550 + 1,600 + 1,650 + ... 4,000

t_n = 1,500 + (n-1)50 S₅₁ = 51(1,500 +4,000)/2

4,000 = 1,500 + 50n - 50 = 51(2,750)

2,550 = 50n = 140,250

51 = n

400

Write in sigma notation: 1 + 1/2 + 1/3 + 1/4 + ... + 1/47

∑ 1/n

n=1

400

Find the 4th term (k=3) for the expansion of

(x²- 3y)¹⁸

18!/(18-3)!3! (x²)^18-3(-3y)³

= 18*17*16*15!/15!*6 (x²)¹⁵(-27y³)

= 3*17*16*(-27) x³⁰y³

= -22,032x³⁰y³

500

A pile of bricks has 85 bricks in the bottom row, 79 bricks in the second row, 73 bricks in the third row, and so on until there is only 1 brick in the top row.

a. How many bricks are in the 12th row?

b. How many rows are there in all?

a. 85 + 79 + 73 + ... + 1 b. 1 = 91 - 6n

t_n = 85 + (n-1)(-6) = 91 - 6n -90 = -6n

t₁₂ = 91-6(12) = 19 bricks n = 15 rows

500

A new pair of running shoes costs $10 now. They double in price every year. How much will they cost in 14 years?

t₁₄= 10*(2)^n-1 = 10*(2)¹⁴^-1 = $81,920

500

Find the sum of:

64/9 + 16/3 + 4 + 3 + ...

S_∞ = (64/9)/(1 - 3/4) r = 3/4

= (64/9)/(1/4)

= 28.4

500

Write in sigma notation:

1/3 + 4/6 + 9/9 + 16/12 + 25/15 + ... 169/39

∑ n²/3n

n=1

500

Find the first 4 terms of the expansion of (x² + 3y)¹⁵

k=0 --> (x²)¹⁵= x³⁰

k=1 --> 15(x²)¹⁴(3y) = 45x²⁸y

k=2 --> 15!/(15-2)!2! (x²)¹³(3y)² = 945x²⁶y²

k=3 --> 15!/(15-3)!3! (x²)¹²(3y)³ = 12,285x²⁴y³

x³⁰ + 45x²⁸y + 945x²⁶y²+ 12,285x²⁴y³