[tex]$\sum_{n=1}(2 n-1)^2$[/tex]
The sum of the series is
Answer (2)
Expand the general term: ( 2 n − 1 ) 2 = 4 n 2 − 4 n + 1 .
Rewrite the sum: ∑ n = 1 N ( 4 n 2 − 4 n + 1 ) = 4 ∑ n = 1 N n 2 − 4 ∑ n = 1 N n + ∑ n = 1 N 1 .
Use the formulas for the sum of squares and integers.
Simplify to get the final answer: 3 N ( 4 N 2 − 1 ) .
Explanation
Understanding the Problem We are asked to find the sum of the series ∑ n = 1 ( 2 n − 1 ) 2 . Since the upper limit of the summation is not specified, we will assume it is a finite number, say N . Therefore, we want to find a closed-form expression for the sum ∑ n = 1 N ( 2 n − 1 ) 2 .
Expanding the General Term First, let's expand the general term of the series: ( 2 n − 1 ) 2 = 4 n 2 − 4 n + 1 .
Rewriting the Sum Now, we can rewrite the sum as follows: n = 1 ∑ N ( 4 n 2 − 4 n + 1 ) = 4 n = 1 ∑ N n 2 − 4 n = 1 ∑ N n + n = 1 ∑ N 1.
Using Known Formulas We will use the formulas for the sum of the first N squares and the sum of the first N integers: n = 1 ∑ N n 2 = 6 N ( N + 1 ) ( 2 N + 1 ) n = 1 ∑ N n = 2 N ( N + 1 ) n = 1 ∑ N 1 = N
Substituting the Formulas Substitute these formulas into the expression: 4 6 N ( N + 1 ) ( 2 N + 1 ) − 4 2 N ( N + 1 ) + N = 3 2 N ( N + 1 ) ( 2 N + 1 ) − 2 N ( N + 1 ) + N
Simplifying the Expression Now, let's simplify the expression: 3 2 N ( N + 1 ) ( 2 N + 1 ) − 2 N ( N + 1 ) + N = 3 2 N ( N + 1 ) ( 2 N + 1 ) − 6 N ( N + 1 ) + 3 N = 3 N [ 2 ( N + 1 ) ( 2 N + 1 ) − 6 ( N + 1 ) + 3 ] = 3 N [ 2 ( 2 N 2 + 3 N + 1 ) − 6 N − 6 + 3 ] = 3 N [ 4 N 2 + 6 N + 2 − 6 N − 3 ] = 3 N ( 4 N 2 − 1 )
Final Answer Therefore, the sum of the series is: n = 1 ∑ N ( 2 n − 1 ) 2 = 3 N ( 4 N 2 − 1 )
Examples
This formula can be used to calculate the total amount of material needed to build a staircase with N steps, where each step's area corresponds to the square of an odd number. For instance, if you are designing a staircase with 5 steps, you can use this formula to quickly determine the total area of the steps. This kind of calculation is useful in construction, architecture, and even in optimizing resource allocation in manufacturing processes.
To find the sum of the series ∑ n = 1 N ( 2 n − 1 ) 2 , we expand the term to 4 n 2 − 4 n + 1 and rewrite the sum using known formulas for summations. This leads us to the final result: ∑ n = 1 N ( 2 n − 1 ) 2 = 3 N ( 4 N 2 − 1 ) .
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