Math Problem Statement

Calculate \( \sum_{n=1}^{20} (3 + 5n) \). By how much would the sum change if \( n \) started at 0 and ended at 19?

Solution

To solve this problem:

Given:

n=120(3+5n)\sum_{n=1}^{20} (3 + 5n) We are asked:

  1. How much the sum would change if nn starts at 0 and ends at 19.

Step 1: Calculate the original sum from n=1n = 1 to n=20n = 20

The summation expression is 3+5n3 + 5n. To compute the sum:

  1. Split the terms: n=120(3+5n)=n=1203+n=1205n\sum_{n=1}^{20} (3 + 5n) = \sum_{n=1}^{20} 3 + \sum_{n=1}^{20} 5n

  2. Calculate each part:

    • First part: n=1203\sum_{n=1}^{20} 3
      Since 3 is a constant added 20 times: n=1203=320=60\sum_{n=1}^{20} 3 = 3 \cdot 20 = 60

    • Second part: n=1205n\sum_{n=1}^{20} 5n
      Factor out the 5: n=1205n=5n=120n\sum_{n=1}^{20} 5n = 5 \cdot \sum_{n=1}^{20} n Use the formula for the sum of the first NN natural numbers: n=120n=20(20+1)2=20212=210\sum_{n=1}^{20} n = \frac{20(20+1)}{2} = \frac{20 \cdot 21}{2} = 210 Thus: n=1205n=5210=1050\sum_{n=1}^{20} 5n = 5 \cdot 210 = 1050

  3. Combine the results: n=120(3+5n)=60+1050=1110\sum_{n=1}^{20} (3 + 5n) = 60 + 1050 = 1110


Step 2: Recalculate the sum from n=0n = 0 to n=19n = 19

Now, nn starts at 0 and ends at 19. The same expression applies:

  1. Split the terms: n=019(3+5n)=n=0193+n=0195n\sum_{n=0}^{19} (3 + 5n) = \sum_{n=0}^{19} 3 + \sum_{n=0}^{19} 5n

  2. Calculate each part:

    • First part: n=0193\sum_{n=0}^{19} 3
      Since 3 is added 20 times (including n=0n = 0): n=0193=320=60\sum_{n=0}^{19} 3 = 3 \cdot 20 = 60

    • Second part: n=0195n\sum_{n=0}^{19} 5n
      Factor out the 5: n=0195n=5n=019n\sum_{n=0}^{19} 5n = 5 \cdot \sum_{n=0}^{19} n Use the formula for the sum of the first NN natural numbers, where N=19N = 19: n=019n=19(19+1)2=19202=190\sum_{n=0}^{19} n = \frac{19(19+1)}{2} = \frac{19 \cdot 20}{2} = 190 Thus: n=0195n=5190=950\sum_{n=0}^{19} 5n = 5 \cdot 190 = 950

  3. Combine the results: n=019(3+5n)=60+950=1010\sum_{n=0}^{19} (3 + 5n) = 60 + 950 = 1010


Step 3: Determine the change in the sum

The original sum (from n=1n = 1 to n=20n = 20) is 1110.
The modified sum (from n=0n = 0 to n=19n = 19) is 1010.

The change in the sum is: 11101010=1001110 - 1010 = 100


Final Answer:

  • The sum decreases by 100.

Do you want further clarification or a detailed breakdown of any part?

Here are 5 related questions you might find useful:

  1. What is the formula for summing an arithmetic series?
  2. How do you calculate the sum of the first NN natural numbers?
  3. How does changing the bounds of a summation affect its result?
  4. What happens when constants are added inside summation notation?
  5. How can you simplify a summation involving both constants and linear terms?

Tip: Always split summations into simpler parts when dealing with sums of constants and variables!

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Math Problem Analysis

Mathematical Concepts

Summation Notation
Arithmetic Series
Linear Equations

Formulas

Sum of first N natural numbers: \( \sum_{k=1}^N k = \frac{N(N+1)}{2} \)
Sum of a constant repeated: \( \sum_{k=1}^N c = c \cdot N \)

Theorems

Properties of Summation

Suitable Grade Level

High School (Grades 9-11)