$$\large\sum_{j=0}^n (-1)^j {n\choose j}={n\choose 0}-{n\choose 1}+…..+\pm{n\choose n}=0 $$

I'm confused by the last part of the equation $\pm$. it seems imply that the sum would be equal to 0 no matter the $n$ is even or odd ?

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# [Math] Using Binomial Theorem to prove the following

###### Related Question

binomial theorembinomial-coefficientssummation

$$\large\sum_{j=0}^n (-1)^j {n\choose j}={n\choose 0}-{n\choose 1}+…..+\pm{n\choose n}=0 $$

I'm confused by the last part of the equation $\pm$. it seems imply that the sum would be equal to 0 no matter the $n$ is even or odd ?

## Best Answer

HINT$$(1-1)^n = \large\sum_{j=0}^n \binom{n}{j}(1)^{n-j}(-1)^{j} $$