Find 101∑x=02x101−13x210201−3x101+1.
In case you are not well prepared to attack the problem analytically, you could still tackle it algebraically, that is purely allowable and no one will ever say algebraic method is not awesome!
For simplicity's sake, we let xi=i101 and f(x)=2x101−13x210201−3x101+1, we then have:
f(x)=2x101−13x210201−3x101+1
=2(x101)−13(x101)2−3(x101)+1
=2xi−13x2i−3xi+1
Furthermore, if we replace xi by 1−xi, we see that:
f(x)=2xi−13x2i−3xi+1
f(1−xi)=2(1−xi)−13(1−xi)2−3(1−xi)+1
=2−2xi−13x2i−6xi+3−3+3xi+1
=1−2xi3x2i−3xi+1
=−(−1−2xi3x2i−3xi+1)
=−(2xi−13x2i−3xi+1)
=−f(xi)
Thus, what we can conclude by now is that we have proved that f(xi)+f(1−xi)=0
Our next step is to make full use of this piece of valuable data and also xi=i101:
1.
Observe that 1−i101=101−i101, this tells us 1−xi=x101−i
2.
We let
S=101∑x=02x101−13x210201−3x101+1=101∑x=02xi−13x2i−3xi+1=101∑x=0f(xi)−−(∗)
We can also have
S=101∑x=0f(1−xi)−−(∗∗)
3.
By adding the two equations (*) and (**) up yields:
2S=101∑x=0(f(xi)+f(1−xi))
=101∑x=0(f(xi)+f(1−xi))
=101∑x=0(0)
4.
Therefore,
S=101∑x=02x101−13x210201−3x101+1=0.
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