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I was going through the section on Binomials in Mathematics: Analyses and Approaches (IB). The author stated the expansion of $(1+x)^n$ as follows:

$$ (1+x)^n = { {{n} \choose {0}} + {{n} \choose {1}}x + {{n} \choose {2}}x^2 + \;...\; + {{n} \choose {n}}x^n} $$

Then, he said that it could be used to prove the following relationships/equations:

$$ 2^n = { {{n} \choose {0}} + {{n} \choose {1}} + {{n} \choose {2}} + \;...\; + {{n} \choose {n}}} $$

$$ 0^n = { {{n} \choose {0}} - {{n} \choose {1}} + {{n} \choose {2}} - {{n} \choose {3}} + \;...\; + (-1)^n{{n} \choose {n}}} $$

I was able to derive these relationships on my own, but got stuck on the expansion of $4^n$, here is the result, based on the original expansion of $(1+x)^n$. Could someone please explain how to derive this? $$ 4^n = { {{2n+1} \choose {0}} + {{2n+1} \choose {1}} + \;...\; + {{2n+1} \choose {n}}} $$

Mehul Jangir
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1 Answers1

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using $\binom mr=\binom m{m-r}$

$$2S=2\sum_{r=0}^n\binom{2n+1}r =\sum_{r=0}^n \binom{2n+1}r+\sum_{r=0}^n\binom{2n+1}{2n+1-r} $$

Set $2n+1-r=p$ $$\sum_{r=0}^n\binom{2n+1}{2n+1-r}=\sum_{p=n+1}^{2n+1}\binom{2n+1}r $$

Consequently, $$2S=\sum_{r=0}^{2n}\binom{2n+1}r =(1+1)^{2n+1}$$

lab bhattacharjee
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