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# Finite Sums Trivia Quiz

### There are many formulas for finding certain sums. I will give you a formula, and you can use it to find the sum of the given finite series. You will need a calculator. Please do not put commas in your answers. Good luck!

A multiple-choice quiz by rodney_indy. Estimated time: 7 mins.

Author
Time
7 mins
Type
Multiple Choice
Quiz #
279,187
Updated
Jul 23 22
# Qns
10
Difficulty
Tough
Avg Score
6 / 10
Plays
499
Awards
Top 35% Quiz
- -
Question 1 of 10
1. I'll begin with the following formula, which is for the sum of the first n positive integers:

1 + 2 + 3 + ... + n = n(n + 1)/2

Using this formula, find the following sum:

1 + 2 + 3 + ... + 74

Answer: (Substitute a certain value for n in the formula)

#### NEXT>

Question 2 of 10
2. Once again, we'll be using the formula for the sum of the first n positive integers:

1 + 2 + 3 + ... + n = n(n + 1)/2

This time, I want you to find the following sum:

75 + 76 + 77 + ... + 124

As a hint, note that the sum above can be found the following way:

(1 + 2 + 3 + ... + 124) - (1 + 2 + 3 + ... + 74)

#### NEXT>

Question 3 of 10
3. Here's another classic formula: The sum of the first n positive odd integers is n^2:

1 + 3 + 5 + ... + (2n - 1) = n^2

Note that the nth positive integer is 2n - 1. Use the above formula to find the following sum:

1 + 3 + 5 + ... + 77

As a hint, you will first need to find the value of n that corresponds to 77.

Answer: (First find n by solving 2n - 1 = 77)

#### NEXT>

Question 4 of 10
4. As above, the sum of the first n positive odd integers is given by n^2:

1 + 3 + 5 + ... + (2n - 1) = n^2.

Use this result to find the following sum:

37 + 39 + 41 + ... + 193

As a hint, get the answer by subtracting two quantities as in question 2.

#### NEXT>

Question 5 of 10
5. Here's another formula:

1 + 4 + 7 + 10 + ... + (3n - 2) = n(3n - 1)/2

Use this one to find the following sum:

1 + 4 + 7 + 10 + ... + 1999

Answer: (First find n by solving 3n - 2 = 1999)

#### NEXT>

Question 6 of 10
6. All of the above formulas were for sums of arithmetic progressions. In each of the sums above, the numbers all differed by the same amount. The remaining ones will be different. The following is the formula for the sum of the squares of the first n positive integers:

1^2 + 2^2 + 3^2 + ... + n^2 = n(n + 1)(2n + 1)/6

Use this formula to find the following sum:

1^2 + 2^2 + 3^2 + ... + 42^2

Answer: (Substitute a certain value for n in the formula)

#### NEXT>

Question 7 of 10
7. The following formula gives the sum of the cubes of the first n positive integers:

1^3 + 2^3 + 3^3 + ... + n^3 = (n(n + 1)/2)^2

Observe that it is the square of the sum of the first n positive integers! Use this formula to find the following sum:

1^3 + 2^3 + 3^3 + ... + 57^3

Answer: (Substitute a certain value for n in the formula)

#### NEXT>

Question 8 of 10
8. Here's a different type of formula:

1*2 + 2*3 + 3*4 + ... + n(n + 1) = n(n + 1)(n + 2)/3

Use this formula to find the following sum:

1*2 + 2*3 + 3*4 + ... + 83*84

Answer: (Substitute a certain value for n in the formula)

#### NEXT>

Question 9 of 10
9. Here's another formula similar to the last one:

1*3 + 2*4 + 3*5 + ... + n(n + 2) = n(n + 1)(2n + 7)/6

Use this formula to find the following sum:

3 + 8 + 15 + ... + 728

As a hint, you first need to find the value of n that satisfies n(n + 2) = 728.

#### NEXT>

Question 10 of 10
10. The last formula is one with a nice sum, but here we are summing a sequence whose general term is a quadratic:

1 + 7 + 19 + 37 + ... + (3n^2 - 3n + 1) = n^3

Use the above formula to find the following sum:

1 + 7 + 19 + 37 + ... + 1951

As a hint, find the value of n first by solving a quadratic equation.

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Quiz Answer Key and Fun Facts
1. I'll begin with the following formula, which is for the sum of the first n positive integers: 1 + 2 + 3 + ... + n = n(n + 1)/2 Using this formula, find the following sum: 1 + 2 + 3 + ... + 74

Here n = 74, so the sum is given by 74*75/2 = 2775.
2. Once again, we'll be using the formula for the sum of the first n positive integers: 1 + 2 + 3 + ... + n = n(n + 1)/2 This time, I want you to find the following sum: 75 + 76 + 77 + ... + 124 As a hint, note that the sum above can be found the following way: (1 + 2 + 3 + ... + 124) - (1 + 2 + 3 + ... + 74)

By the formula,

1 + 2 + 3 + ... + 124 = 124*125/2 = 7750,

1 + 2 + 3 + ... + 74 = 2775 (from question 1)

Therefore, 75 + 76 + 77 + ... + 124 = 7750 - 2775 = 4975.
3. Here's another classic formula: The sum of the first n positive odd integers is n^2: 1 + 3 + 5 + ... + (2n - 1) = n^2 Note that the nth positive integer is 2n - 1. Use the above formula to find the following sum: 1 + 3 + 5 + ... + 77 As a hint, you will first need to find the value of n that corresponds to 77.

The nth term is given by 2n - 1. We need to first find the value of n that corresponds to an nth term of 77, so set 2n - 1 equal to 77 and solve for n:

2n - 1 = 77

2n = 77 + 1

2n = 78

n = 39

By the formula, the sum is n^2. Hence the sum is 39^2 = 1521.
4. As above, the sum of the first n positive odd integers is given by n^2: 1 + 3 + 5 + ... + (2n - 1) = n^2. Use this result to find the following sum: 37 + 39 + 41 + ... + 193 As a hint, get the answer by subtracting two quantities as in question 2.

First note that

37 + 39 + 41 + ... + 193 = (1 + 3 + 5 + ... + 193) - (1 + 3 + 5 + ... + 35)

Now find the term that corresponds to 193:

2n - 1 = 193

2n = 194

n = 194/2 = 97

So using the formula, 1 + 3 + 5 + ... + 193 = 97^2 = 9409.

Now find the term that corresponds to 35:

2n - 1 = 35

2n = 36

n = 36/2 = 18

So using the formula, 1 + 3 + 5 + ... + 35 = 18^2 = 324.

Therefore, 37 + 39 + 41 + ... + 193 = 9409 - 324 = 9085.
5. Here's another formula: 1 + 4 + 7 + 10 + ... + (3n - 2) = n(3n - 1)/2 Use this one to find the following sum: 1 + 4 + 7 + 10 + ... + 1999

First find the term corresponding to 1000 by solving 3n - 2 = 1999:

3n = 2001

n = 2001/3 = 667

So by the formula,

1 + 4 + 7 + 10 + ... + 1999 = 667(3*667 - 1)/2 = 667*2000/2 = 667000
6. All of the above formulas were for sums of arithmetic progressions. In each of the sums above, the numbers all differed by the same amount. The remaining ones will be different. The following is the formula for the sum of the squares of the first n positive integers: 1^2 + 2^2 + 3^2 + ... + n^2 = n(n + 1)(2n + 1)/6 Use this formula to find the following sum: 1^2 + 2^2 + 3^2 + ... + 42^2

Substitute n = 42 in the formula:

1^2 + 2^2 + 3^2 + ... + 42^2 = 42*43*(2*42 + 1)/6 = 42*43*85/6 = 25585.

The formula for the sum of the first n squares appears in calculus books. It is used in the process of computing certain definite integrals (such as the integral of x^2 dx on the interval [0,1]) by taking the limit of a certain Riemann sum.
7. The following formula gives the sum of the cubes of the first n positive integers: 1^3 + 2^3 + 3^3 + ... + n^3 = (n(n + 1)/2)^2 Observe that it is the square of the sum of the first n positive integers! Use this formula to find the following sum: 1^3 + 2^3 + 3^3 + ... + 57^3

Put n = 57 in the formula:

1^3 + 2^3 + 3^3 + ... + 57^3 = (57*58/2)^2 = 1653^2 = 2732409.

This formula is also used in calculus books. Fortunately, once you learn the fundamental theorem of calculus you will no longer need to use these formulas to compute definite integrals of polynomials.
8. Here's a different type of formula: 1*2 + 2*3 + 3*4 + ... + n(n + 1) = n(n + 1)(n + 2)/3 Use this formula to find the following sum: 1*2 + 2*3 + 3*4 + ... + 83*84

Here n = 83:

1*2 + 2*3 + 3*4 + ... + 83*84 = 83*84*85/3 = 197540.

Formulas such as this one and all the other ones in this quiz can be proven by a procedure called mathematical induction. But where did it come from in the first place? This formula probably has an "elegant" derivation, but here is how I derived it. If you are comfortable with sigma notation, here is how you do it. I will use "Sum" for a capital sigma. We want a formula for
Sum(i = 1 to n, i(i + 2)).

Expand: Sum(i = 1 to n, i(i + 2)) = Sum(i = 1 to n, i^2 + 2i)

= Sum(i = 1 to n, i^2) + 2 * Sum(i = 1 to n, i) [summation is linear]

= n(n + 1)(2n + 1)/6 + 2 * n(n + 1)/2 [use previous formulas]

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

= n(n + 1)/6 * 2(n + 2)

= n(n + 1)(n + 2)/3
9. Here's another formula similar to the last one: 1*3 + 2*4 + 3*5 + ... + n(n + 2) = n(n + 1)(2n + 7)/6 Use this formula to find the following sum: 3 + 8 + 15 + ... + 728 As a hint, you first need to find the value of n that satisfies n(n + 2) = 728.

First we need to find the term that corresponds to 728 by solving n(n + 2) = 728:

n(n + 2) = 728

n^2 + 2n = 728 Completing the square makes this quick here:

n^2 + 2n + 1 = 729

(n + 1)^2 = 27^2

Since n is positive, we have n + 1 = 27, thus n = 26. Substitute this value of n into the formula:

3 + 8 + 15 + ... + 728 = 26*27*59/6 = 6903

This formula can be derived in a similar way as the formula in question 8.
10. The last formula is one with a nice sum, but here we are summing a sequence whose general term is a quadratic: 1 + 7 + 19 + 37 + ... + (3n^2 - 3n + 1) = n^3 Use the above formula to find the following sum: 1 + 7 + 19 + 37 + ... + 1951 As a hint, find the value of n first by solving a quadratic equation.

First find n by solving the quadratic equation 3n^2 - 3n + 1 = 1951:

3n^2 - 3n + 1 - 1951 = 0

3n^2 - 3n - 1950 = 0

3(n^2 - n - 650) = 0

3(n + 25)(n - 26) = 0

Setting factors to zero gives n = 26. Substitute this value of n into the formula to get the answer:

1 + 7 + 19 + 37 + ... + 1951 = 26^3 = 17576.

I hope you enjoyed this quiz! Thanks for playing!
Source: Author rodney_indy

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