Question

Find $P_1 + P_2 + \dots + P_{10}$ if $P_k$ is the perimeter of a square having side length $k$. 
 

Hint:

Whenever perimeter depends linearly on $k$, factor out constants first and then use the formula for sum of natural numbers.

Answer 1

Correct Answer: 220

Step 1: Understanding the Question
The problem asks for the sum of the perimeters of 10 squares. The side length of the first square is 1, the second is 2, and so on, up to the tenth square with a side length of 10.
Step 2: Key Formula or Approach
First, we need the formula for the perimeter of a square. For a square with side length 's', the perimeter 'P' is: \[ P = 4s \] Second, we need the formula for the sum of the first 'n' natural numbers (an arithmetic series): \[ S_n = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} \] Step 3: Detailed Explanation
First, let's express the perimeter $P_k$ for a square with side length $k$. Using the perimeter formula: \[ P_k = 4k \] Now, we need to find the sum $S = P_1 + P_2 + \dots + P_{10}$. Substituting the expression for $P_k$: \[ S = 4(1) + 4(2) + 4(3) + \dots + 4(10) \] We can factor the constant '4' out of the sum: \[ S = 4 \times (1 + 2 + 3 + \dots + 10) \] The expression in the parentheses is the sum of the first 10 natural numbers. We can use the arithmetic series formula with $n=10$: \[ 1 + 2 + \dots + 10 = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = \frac{110}{2} = 55 \] Finally, we multiply this result by the factor of 4 that we took out earlier: \[ S = 4 \times 55 = 220 \] Step 4: Final Answer
The sum of the perimeters is 220.
\[ \boxed{220} \]