Evaluate the following $(3-2)(5-3)(7-5)........(1995-1993)$

Question

Evaluate the following $(3-2)(5-3)(7-5)....(1995-1993)$

I have problem in counting the terms since the answer will be $2^n$ , $n$ is the number of terms.

So I count them as follows:

$1995=2n-1$

$1996=2n \to\ n= 998$ ; but since the first term equals $1$ The the answer will be $2^{997}$

Is my answer right?

Answer 1

Since the first factor is indeed $3-2=1$, we will simply ignore it and start by counting $5-3$ as factor #1.

Factor $\#\color{red}{1}$ is $5-3$, which is $(2 \cdot \color{red}{2} + 1) - (2 \cdot \color{red}{2} - 1)$.

Factor $\#\color{red}{2}$ is $7-5$, which is $(2 \cdot \color{red}{3} + 1) - (2 \cdot \color{red}{3} - 1)$.

Factor $\#\color{red}{3}$ is $9-7$, which is $(2 \cdot \color{red}{4} + 1) - (2 \cdot \color{red}{4} - 1)$.

See the pattern?

Factor $\#\color{red}{n}$ is $(2 \cdot \color{red}{[n+1]} + 1) - (2 \cdot \color{red}{[n+1]} - 1)$.

Which $n$ gives us $2[n+1] + 1 = 1995$ and $2[n+1] - 1 = 1993$? It's the same $n$ for both so we only need to solve one of them. Either way we get $n = 996$.

So yes, there are $997$ total factors. One of them is $1$ and the other $996$ are $2$, so the answer is $1 \cdot 2^{996} = 2^{996}$.

Answer 2

Not only is the first factor (not term) equal to $1$, it also starts with a $3$ instead of a $1$ (or the second one with a $5$, if you wish).

The first factor is $1$ so you can omit that factor in your calculation; you then have: $$(5-3)(7-5)\ldots(1995-1993)$$ Since $1995=5+2n \iff n=995$, you have $995$ factors after the first one, since the first one corresponds to $n=0$ giving you $996$ factors of $2$ in total. This means it simplifies to: $$1\cdot 2^{996} = 2^{996}$$

Or more schematically, the first term of all factors after the first one are of the form $\color{blue}{5+2n}$ with $n$ ranging from $0$ to $995$ since $1995=5+2 \cdot 995$:

$$(3-2)\underbrace{(\color{blue}{5}-3)}_{n=0}\underbrace{(\color{blue}{7}-5)}_{n=1}\underbrace{(\color{blue}{9}-7)}_{n=2}\ldots\underbrace{(\color{blue}{1995}-1993)}_{n=995}$$

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Alternatively: include the first factor and note they are also all of the form $\color{red}{3+2n}$ with $n$ ranging from $0$ to $996$, you then have:

$$\underbrace{(\color{red}{3}-2)}_{n=0}\underbrace{(\color{red}{5}-3)}_{n=1}\underbrace{(\color{red}{7}-5)}_{n=2}\underbrace{(\color{red}{9}-7)}_{n=3}\ldots\underbrace{(\color{red}{1995}-1993)}_{n=996}$$ This works out nicely since the factor corresponding to $n=0$ is precisely the factor $3-2=1$ and thus not contributing to the product; all the other factors corresponding to the $n$'s ranging from $1$ to $996$ are all $2$, so...

Answer 3

Your error is that if you are writing $1995=2n-1$, then you need to write the $5$ in $5-3$ in the same form. So the terms $(2n-1-(2n-3))$ starts at $n=3$ and ends at $n=998$. That means you removed two terms, $n=1,n=2$. So the answer is $2^{996}$.

Answer 4

Considering the first two or three terms, $(3-2)(5-3)$ or $(3-2)(5-3)(7-5)$, it seems that you just need to follow

$5=2n+3\rightarrow n=1$ or $7=2n+3\rightarrow n=2$

and finally $1995=2n+3\rightarrow n=996$.