So $ E(n) = \sum_{k=0}^{n-1} (2k + 1) = n^2 $, since sum of first $ n $ odd numbers is $ n^2 $.
["# Why $ \sum_{k=0}^{n-1} (2k + 1) = n^2 $: The Simple Mathematics Behind the Formula", "Mathematics teaches us that some patterns are as elegant as they are powerful. One of the most striking is the identity that the sum of the first $ n $ odd numbers equals $ n^2 $. This elegant equation—$ \sum_{k=0}^{n-1} (2k + 1) = n^2 $—not only demonstrates a beautiful algebraic truth but also connects geometry, number theory, and computation in a remarkable way.", "## What Is the Sum of the First $ n $ Odd Numbers?", "The odd numbers begin at 1 and follow the sequence:\n$ 1, 3, 5, 7, \dots $", "When we write the first $ n $ odd numbers in summation form, they follow a general pattern:\n$$\n\sum_{k=0}^{n-1} (2k + 1)\n$$\nThis expression starts at $ k = 0 $:\n- When $ k = 0 $: $ 2(0) + 1 = 1 $\n- When $ k = 1 $: $ 2(1) + 1 = 3 $\n- When $ k = 2 $: $ 2(2) + 1 = 5 $\n- …\n- When $ k = n-1 $: $ 2(n-1) + 1 = 2n - 1 $", "So the sum adds up all odd numbers from 1 to $ 2n - 1 $.", "## Why Does the Sum Equal $ n^2 $?", "### Pattern Recognition Through Sum", "Let’s compute the sum directly for small $ n $ to spot the pattern:", "- For $ n = 1 $:\n $$\n \sum_{k=0}^{0} (2k + 1) = 1 = 1^2\n $$", "- For $ n = 2 $:\n $$\n \sum_{k=0}^{1} (2k + 1) = 1 + 3 = 4 = 2^2\n $$", "- For $ n = 3 $:\n $$\n 1 + 3 + 5 = 9 = 3^2\n $$", "- For $ n = 4 $:\n $$\n 1 + 3 + 5 + 7 = 16 = 4^2\n $$", "Clearly, the total grows perfectly as the square of $ n $. But why?", "### Pairing and Geometric Insight", "A geometric interpretation helps unveil the truth: imagine forming a square using unit blocks.", "- Start with a single block (a $ 1 \ imes 1 $ square).\n- Add a ring of 3 blocks around it → forms a $ 3 \ imes 3 $ square ($ 3^2 = 9 $).\n- Add another ring of 5 blocks → forms a $ 5 \ imes 5 $ square ($ 5^2 = 25 $).\n- Continue this: each new ring adds $ 2k + 1 $ squares, building outward layer by layer.", "Each "ring" corresponds to the $ k $-th odd number: at layer $ k $, $ 2k + 1 $ blocks are added. The cumulative total after $ n $ layers is exactly the area of an $ n \ imes n $ square:\n$$\nn^2\n$$", "### Proof by Induction", "For a rigorous explanation, we can use mathematical induction.", "Base Case ($ n = 1 $):\n$$\n\sum_{k=0}^{0} (2k + 1) = 1 = 1^2\n$$\nTrue.", "Inductive Step:\nAssume $ \sum_{k=0}^{n-1} (2k + 1) = n^2 $ holds. Now compute for $ n + 1 $:\n$$\n\sum_{k=0}^{n} (2k + 1) = \left( \sum_{k=0}^{n-1} (2k + 1) \right) + (2n + 1)\n= n^2 + 2n + 1 = (n + 1)^2\n$$\nThus, the identity holds for $ n + 1 $. By induction, it holds for all $ n \geq 1 $.", "## Practical Applications", "Beyond pure math, this identity is useful in:\n- Computer Science: Efficient algorithms for prefix sums of odd numbers.\n- Education: Teaching pattern recognition and proof techniques.\n- Engineering: Modeling structured growth, like layer-by-layer construction.", "## Conclusion", "The formula $ \sum_{k=0}^{n-1} (2k + 1) = n^2 $ is a brilliant example of how simple arithmetic unveils deep mathematical truths. It connects abstract formulas to tangible geometric shapes, supports inductive reasoning, and offers practical utility. Whether you're a student, teacher, or enthusiast, understanding this identity deepens your appreciation for the harmony of numbers and structure in mathematics.", "---", "Keywords: $ \sum_{k=0}^{n-1} (2k + 1) $, $ n^2 $ identity, sum of odd numbers, mathematical proof, pattern recognition, inductive proof, geometry and math, algorithmic insight, education resource."]
