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?


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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