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Can someone please help me with this? Evaluate the sum (n¦0)-2(n¦1)+4(n¦2)-8(n¦3)+…〖-2〗^n (n¦n) for n=3&4. After I find the sum I need to use the binomial theorem to verify my findings.

So for n=3 I got -1 and for n=4 I got 1. I noticed that when I was solving the (n¦n) part I would get the numbers from Pascal’s triangle for the corresponding rows. Does this have anything to do with the question?

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The ordinary Binomial Theorem can be stated in various equivalent ways. One of them is that if $n$ is a non-negative integer, then $$(1+x)^n =\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\cdots+\binom{n}{n}x^n.$$ The numbers $\binom{n}{k}$ are indeed the entries in the $n$-th row of the so-called Pascal Triangle.

For your problem, set $x=-2$.

Remark: Another version is $$(a+b)^n=\binom{n}{0}a^n +\binom{n}{1}a^{n-1}b+\binom{n}{2}a^{n-2}b^2 +\cdots +\binom{n}{n}b^n.$$ Then set $a=1$, $b=-2$.

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