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Formulate Newton's binomial's solution and, using that, deduce that this formula is true: $${n\choose 0} + {n\choose 1} + \cdots + {n\choose n-1}+{n\choose n} = 2^n$$

I know that Newton's binomial is $$(a+b)^n={n\choose 0}a^n + {n\choose 1}a^{n-1}b +\cdots +{n\choose n-1}ab^{n-1} + {n\choose n}b^n$$

Can anybody help me with the second part? Thank you

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    Welcome to Math.SE. Sorry but I don't understand your equations. Here's a [tutorial](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) on formatting your equations. – 2017-01-04
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    Put $$a=b=1$$ in the identity – 2017-01-04

2 Answers 2

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Hint:

$$2^n=(1+1)^n$$

Now apply binomial theorem on $(1+1)^n$.

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If you know that

$$(a+b)^n={n\choose 0}a^n + {n\choose 1}a^{n-1}b +\cdots +{n\choose n-1}ab^{n-1} + {n\choose n}b^n$$

Then when $a=b=1$, we have

$$(1+1)^n={n\choose 0} + {n\choose 1} + \cdots + {n\choose n-1}+{n\choose n}$$

And it follows from there.