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Distribute the sum:
Use the following formulas:
In case you don't know where those formulas came from:
The first one is obvious; you're just adding <em>n</em> copies of 1, so 1 + 1 + ... + 1 = <em>n</em>.
The second can be proved in this way: let <em>S</em> be the sum 1 + 2 + 3 + ... + <em>n</em>. Rearrange it as <em>S</em> = <em>n</em> + (<em>n</em> - 1) + (<em>n</em> - 2) + ... + 1. Then 2<em>S</em> = (<em>n</em> + 1) + (<em>n</em> + 1) + (<em>n</em> + 1) + ... + (<em>n</em> + 1), or <em>n</em> copies of <em>n</em> + 1. So 2<em>S</em> = <em>n</em>(<em>n</em> + 1). Divide both sides by 2 and we're done.
Answer:
See below
Step-by-step explanation:
Given the sets