Math  /  Algebra

Question210+211+212++220042n2m2^{10}+2^{11}+2^{12}+\ldots+2^{2004} \rightarrow 2^{n}-2^{m} m,n=m, n= natural numbers

Studdy Solution

STEP 1

What is this asking? We're asked to rewrite the sum of powers of 2, from 2102^{10} all the way up to 220042^{2004}, in the much simpler form of 2n2m2^n - 2^m, where *n* and *m* are natural numbers. Watch out! It's easy to mess up the powers when factoring.
Double-check each step carefully!

STEP 2

1. Factor the expression
2. Simplify using the geometric series formula
3. Express in the desired form

STEP 3

Alright, let's **factor out** the smallest power of 2 from our sum, which is 2102^{10}!
This is like reverse distributing, and it's a super useful trick!

STEP 4

210+211+212++22004=210(1+21+22++21994)2^{10} + 2^{11} + 2^{12} + \ldots + 2^{2004} = 2^{10}(1 + 2^1 + 2^2 + \ldots + 2^{1994}) See how neat that is?
We've got 2102^{10} multiplied by a much simpler sum inside the parentheses.

STEP 5

Now, peep this: the sum inside the parentheses 1+21+22++219941 + 2^1 + 2^2 + \ldots + 2^{1994} is a **geometric series**!
And we've got a handy-dandy formula for that.
Remember, a geometric series is a sum where each term is multiplied by the same value (the **common ratio**) to get the next term.
Here, our common ratio is **2**.

STEP 6

The formula for the sum of a geometric series is a(rn1)r1\frac{a(r^n - 1)}{r - 1}, where *a* is the **first term** (here, a=1a = 1), *r* is the **common ratio** (here, r=2r = 2), and *n* is the **number of terms**.
How many terms do we have?
Well, the powers of 2 go from 0 to 1994, so we have 19940+1=19951994 - 0 + 1 = \textbf{1995} terms.

STEP 7

Plugging these values into our formula, we get: 1+21+22++21994=1(219951)21=2199511=2199511 + 2^1 + 2^2 + \ldots + 2^{1994} = \frac{1 \cdot (2^{1995} - 1)}{2 - 1} = \frac{2^{1995} - 1}{1} = 2^{1995} - 1 So, our original expression becomes: 210(219951)2^{10}(2^{1995} - 1)

STEP 8

Let's **distribute** that 2102^{10} back in: 210(219951)=210219952101=210+1995210=220052102^{10}(2^{1995} - 1) = 2^{10} \cdot 2^{1995} - 2^{10} \cdot 1 = 2^{10 + 1995} - 2^{10} = 2^{2005} - 2^{10}

STEP 9

Boom! We've got it in the form 2n2m2^n - 2^m, where n=2005n = \textbf{2005} and m=10m = \textbf{10}.

STEP 10

m=10m = 10 and n=2005n = 2005

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