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By considering powers of (1+x), show that the sum of the squares of the binomial coefficients from 0 to n is 2nCn

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What are the possible remainders when the 100-th power of an integer is divided by 125?

Summit

Stage: 5 Challenge Level: Challenge Level:1
This neat solution came from Marcos and the result was also proved by Yatir Halevi:
By the binomial expansion:

$$(1+x)^m=\sum_{t=0}^m \frac{m!}{t!(m-t)!}x^t $$

This can be proved by induction on $m$ but I won't clutter this with unnecessary proofs.

Putting in $x= -1$ we have

$$0=\sum_{t=0}^m \frac{m!}{t!(m-t)!}(-1)^t $$

Dividing through by $m!$ gives us the required result:

$$\sum_{t=0}^m \frac{(-1)^t}{t!(m-t)!}=0 $$