It starts with Negative?

Algebra Level 2

5 ∑ n = − 2 ∞ 1 10 ⋅ 2 n = ? \large 5 \sum_{n=-2}^\infty \frac 1 {10 \cdot 2^n} = \ ?


The answer is 4.

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Victor Paes Plinio by Victor Paes Plinio , 21, Brazil

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5 solutions

Chew-Seong Cheong
Mar 27, 2015

5 ∑ n = − 2 ∞ 1 10 ˙ 2 n = ∑ n = − 2 ∞ 1 2 ˙ 2 n = ∑ n = − 2 ∞ 1 2 ( n + 1 ) = ∑ n = − 1 ∞ 1 2 n = 2 + ∑ n = 0 ∞ 1 2 n = 2 + 1 1 − 1 2 = 2 + 2 = 4 \displaystyle 5\sum_{n=-2}^\infty {\dfrac {1}{10\dot{}2^n}} = \sum_{n=-2}^\infty {\dfrac {1}{2\dot{}2^n}} = \sum_{n=-2}^\infty {\dfrac {1}{2^{(n+1)}}} = \sum_{n=-1}^\infty {\dfrac {1}{2^n}} = 2+\sum_{n=0}^\infty {\dfrac {1}{2^n}} \\ \quad \quad \quad \quad \quad \quad = 2 + \dfrac{1}{1-\frac{1}{2}} = 2+2=\boxed{4}

15 Helpful 0 Interesting 0 Brilliant 0 Confused

You can clearly factor out 1 10 \frac{1}{10} and proceed to an infinite geometric sum.

so that is, 5 ∑ n = − 2 ∞ 1 10 ⋅ 2 n \displaystyle\large 5 \sum_{n=-2}^\infty \frac 1{10 \cdot 2^n}

= 5 10 ∑ n = − 2 ∞ 1 2 n = 1 2 ⋅ 1 2 − 2 1 − 1 2 = 1 2 ⋅ 4 ⋅ 2 = 4 \displaystyle=\frac 5{10} \sum_{n=-2}^\infty \frac1{2^n} = \frac1{2} \cdot \frac{\frac{1}{2^{-2}}}{1-\frac1{2}} = \frac1{2} \cdot 4 \cdot 2 = \boxed{4}

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Lucas Mayol
Apr 1, 2015

Another solution :

5 ∑ n = − 2 ∞ 1 10 ⋅ 2 n = ∑ n = 0 ∞ 5 10 ⋅ 2 n + 5 ∑ n = − 2 − 1 1 10 ⋅ 2 n \displaystyle 5 \sum_{n=-2}^\infty \frac 1 {10 \cdot 2^n} = \sum_{n=0}^\infty \frac 5 {10 \cdot 2^n} + 5 \sum_{n=-2}^{-1} \frac 1 {10 \cdot 2^n}

Ok so:

∑ n = 0 ∞ 5 10 ⋅ 2 n = ∑ n = 0 ∞ 5 5 ⋅ 2 ⋅ 2 n = ∑ n = 0 ∞ 1 2 ⋅ 2 n = ∑ n = 1 ∞ 1 2 n = 1 2 + 1 4 + 1 8 + . . . = 1 \displaystyle \sum_{n=0}^\infty \frac 5 {10 \cdot 2^n} = \sum_{n=0}^\infty \frac 5 {5 \cdot 2 \cdot 2^n} = \sum_{n=0}^\infty \frac 1 { 2 \cdot 2^n} = \sum_{n=1}^\infty \frac 1 {2^n} = \frac 1 {2} + \frac 1 {4} + \frac 1 {8} + . . . = 1

wikipedia

And now:

5 ∑ n = − 2 − 1 1 10 ⋅ 2 n = 5 ( 2 10 + 4 10 ) = 30 10 = 3 \displaystyle 5 \sum_{n=-2}^{-1} \frac 1 {10 \cdot 2^n} =5 (\frac 2 {10} + \frac 4 {10}) = \frac {30} {10} = 3

Result :

5 ∑ n = − 2 ∞ 1 10 ⋅ 2 n = 3 + 1 = 4 \displaystyle 5 \sum_{n=-2}^\infty \frac 1 {10 \cdot 2^n} = 3 + 1 = 4

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Cs ಠ_ಠ Lee
Mar 29, 2015

Easy

multiply 5 by 1/(10*2^n) to get 1/2^n+1

Since n ranges from -2 and goes to infinity,

the sequence goes like so:

2, 1, 1/2, 1/4....

2+1+1/2+1/4...

the 1/2+1/4.... part (using the telescoping method)

x = 1/2 + 1/4 + 1/8... 2x = 1+ 1/2 + 1/4... 2x = 1 + x x= 1

2+1+1= 4

1 Helpful 0 Interesting 0 Brilliant 0 Confused

5 ∑ n = − 2 ∞ 1 10 × 2 n = 1 2 ∑ n = − 2 ∞ 1 2 n = 1 2 ( 1 2 − 2 + 1 2 − 1 + 1 2 0 + 1 2 1 + 1 2 2 + . . . ) 5\sum\limits_{n=-2}^\infty \frac{1}{10\times2^{n}} =\frac{1}{2}\sum\limits_{n=-2}^\infty \frac{1}{2^{n}} =\frac{1}{2}(\frac{1}{2^{-2}}+\frac{1}{2^{-1}}+\frac{1}{2^{0}}+\frac{1}{2^{1}}+\frac{1}{2^{2}}+...)

The expression inside the parentheses can be interpreted as the sum of an infinite geometric sequence with the first term 4 4 and a common ratio of 1 2 \frac{1}{2} .

Applying the formula for the sum of an infinite geometric sequence, 1 2 ( 4 1 − 1 2 ) = 4 \frac{1}{2}(\frac{4}{1-\frac{1}{2}})=\boxed{4}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Same as I did.

Manish Mayank - 6 years, 2 months ago

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