Thread: Limit and two numbers
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Old July 5th, 2009, 12:15 AM
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Quote:
Originally Posted by Prove It View Post
You need to use the rule a^n - b^n = (a - b)(a^{n - 1} + a^{n - 2}b + a^{n - 3}b^2 + \dots + a^2b^{n - 3} + ab^{n - 2} + b^{n - 1})


You should find something cancels and you're left with a nice easy sum...
Hello: you can let : x-1=t
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