You can put this solution on YOUR website!
Note that the signs alternate: plus then minus then plus .... This part of the pattern can be achieved using consecutive powers of -1.
Note also that all but the first term are powers of x. Looking at the exponents for these powers of x, they start at 1 and go up, one by one, to 2k. So it would seem that a power of x will be part of the general term we will be using. But what about the first term, the 1? If you look at the exponents in reverse order you might get the answer. The exponents of the powers of x, from highest to lowest, starts with 2k and works its way down to 1. So what would the "next" exponent be? Zero? Can we use x to the zero power here? The answer: Yes! so we can rewrite the series as:
We are now close to our final answer. We know that the general term is a power of -1 times a power of x. We just have to figure out exactly which powers of -1 and x and what the limit is for n.
Let's look at the powers of x first. We want the first power of x to be 0. But the n starts at 1. So we need to use (n-1) as the exponent for x.
The limit for n (which goes on top of the sigma) represents the highest possible value for n. We want this highest possible n to make our exponent for x be 2k. Since we have already decided that the general exponent for the x is (n-1), then n-1 = 2k for the highest n. This tells use that the limit for n must be 2k+1.
Finally, we will find the exponent for -1. We want the first term to be positive so when n = 1 we want the exponent for -1 to be even. So we could use (n-1) or (n+1) or (n plus any odd number) for the exponent for -1.
Putting all this together...
can be rewritten as:
By choosing the exponent for -1 that "accidentally" matches the exponent for x, we can condense this down to:
P.S. Remember that we could have chosen a lot of different exponents for -1. So the answer above is not the only one possible.
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