Number of factors is too easy

Number Theory Level 4

1000 ∑ n = 0 ∞ σ 0 ( 6 n ) 6 n 1000\sum _{ n=0 }^{ \infty }{ \frac { { \sigma }_{ 0 }\left( { 6 }^{ n } \right) }{ { 6 }^{ n } } }

Find the value of the expression above if σ 0 ( x ) { \sigma }_{ 0 }\left( x \right) is the number of factors of x x . It is also called the divisor function.

As an explicit example, 4 has 3 factors 1,2 and 4, so σ 0 ( 4 ) = 3 { \sigma }_{ 0 }\left( 4 \right)=3 .


The answer is 2016.

View wiki
Joel Yip by Joel Yip , 21, Singapore

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Joel Yip
Mar 16, 2016

You might want to see this first

∑ n = 0 ∞ σ 0 ( 6 n ) 6 n = ∑ n = 0 ∞ σ 0 ( 2 n × 3 n ) 6 n = ∑ n = 0 ∞ ( n + 1 ) ( n + 1 ) 6 n s i n c e 2 a n d 3 a r e p r i m e . = ∑ n = 0 ∞ ( n + 1 ) 2 6 n \sum _{ n=0 }^{ \infty }{ \frac { { \sigma }_{ 0 }\left( { 6 }^{ n } \right) }{ { 6 }^{ n } } } =\sum _{ n=0 }^{ \infty }{ \frac { { \sigma }_{ 0 }\left( { 2 }^{ n }\times { 3 }^{ n } \right) }{ { 6 }^{ n } } } \\ =\sum _{ n=0 }^{ \infty }{ \frac { \left( n+1 \right) \left( n+1 \right) }{ { 6 }^{ n } } } \quad \quad \quad \quad \quad since\quad 2\quad and\quad 3\quad are\quad prime.\\ =\sum _{ n=0 }^{ \infty }{ \frac { { \left( n+1 \right) }^{ 2 } }{ { 6 }^{ n } } }

∑ n = 0 ∞ x n = 1 1 − x ∀ ∣ x ∣ ≤ 1 \sum_{n=0}^{ \infty } x^{n} = \dfrac{1}{1-x} \quad \forall \quad |x| \leq 1

⇒ ∑ n = 0 ∞ ( d d x x n ) = d d x ( 1 1 − x ) ⇒ ∑ n = 0 ∞ ( n ⋅ x n ) = x ( 1 − x ) 2 ⇒ ∑ n = 0 ∞ d d x ( n ⋅ x n ) = d d x ( x ( 1 − x ) 2 ) ⇒ ∑ n = 0 ∞ ( n 2 ⋅ x n ) = x ⋅ ( 1 − x ) 2 + 2 x ⋅ ( 1 − x ) ( 1 − x ) 4 = x + x 2 ( 1 − x ) 3 \Rightarrow \sum_{n=0}^{ \infty } ( \frac{d}{dx} x^{n} ) = \frac{d}{dx} ( \dfrac{1}{1-x} ) \\ \Rightarrow \sum_{n=0}^{ \infty} ( n \cdot x^{n} ) = \dfrac{x}{ (1-x)^{2} } \\ \Rightarrow \sum_{n=0}^{ \infty} \frac{d}{dx} (n \cdot x^{n} ) = \frac{d}{dx} ( \dfrac{x}{ (1-x)^{2} } ) \\ \Rightarrow \sum_{n=0}^{ \infty} ( n^{2} \cdot x^{n} ) = x\cdot \dfrac{ (1-x)^{2} + 2x\cdot (1-x) }{ (1-x)^{4} } \\= \frac{x+x^{2}}{ (1-x)^{3} }

so now we have

∑ n = 0 ∞ 1 6 n = 1 1 − 1 6 = 6 5 \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { 6 }^{ n } } } =\frac { 1 }{ 1-\frac { 1 }{ 6 } } =\frac { 6 }{ 5 }

∑ n = 0 ∞ n 6 n = 1 6 ( 1 − 1 6 ) 2 = 6 25 \sum _{ n=0 }^{ \infty }{ \frac { n }{ { 6 }^{ n } } } =\frac { \frac { 1 }{ 6 } }{ { \left( 1-\frac { 1 }{ 6 } \right) }^{ 2 } } =\frac { 6 }{ 25 }

∑ n = 0 ∞ n 2 6 n = 1 6 + ( 1 6 ) 2 ( 1 − 1 6 ) 3 = 42 125 \sum _{ n=0 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 6 }^{ n } } } =\frac { { \frac { 1 }{ 6 } +\left( \frac { 1 }{ 6 } \right) }^{ 2 } }{ { \left( 1-\frac { 1 }{ 6 } \right) }^{ 3 } } =\frac { 42 }{ 125 }

∑ n = 0 ∞ ( n + 1 ) 2 6 n = ∑ n = 0 ∞ n 2 + 2 n + 1 6 n = ∑ n = 0 ∞ n 2 6 n + 2 ∑ n = 0 ∞ n 6 n + ∑ n = 0 ∞ 1 6 n = 42 125 + 2 × 6 25 + 6 5 = 252 125 \\ \sum _{ n=0 }^{ \infty }{ \frac { { \left( n+1 \right) }^{ 2 } }{ { 6 }^{ n } } } =\sum _{ n=0 }^{ \infty }{ \frac { { n }^{ 2 }+2n+1 }{ { 6 }^{ n } } } \\ =\sum _{ n=0 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 6 }^{ n } } } +2\sum _{ n=0 }^{ \infty }{ \frac { { n } }{ { 6 }^{ n } } } +\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { 6 }^{ n } } } \\ =\frac { 42 }{ 125 } +2\times \frac { 6 }{ 25 } +\frac { 6 }{ 5 } \\ =\frac { 252 }{ 125 }

OR

∑ n = 0 ∞ ( n + 1 ) 2 6 n = ∑ n = 1 ∞ n 2 6 n − 1 = 6 ∑ n = 1 ∞ n 2 6 n = 6 ( ∑ n = 0 ∞ n 2 6 n − 0 ) = 6 × 42 125 = 252 125 \sum _{ n=0 }^{ \infty } \frac { \left( n+1 \right) ^{ 2 } }{ 6^{ n } } \\ =\sum _{ n=1 }^{ \infty } \frac { n^{ 2 } }{ 6^{ n-1 } } \\ =6\sum _{ n=1 }^{ \infty } \frac { n^{ 2 } }{ 6^{ n } } \\ =6\left( \sum _{ n=0 }^{ \infty } \frac { n^{ 2 } }{ 6^{ n } } -0 \right) \\ =6\times \frac { 42 }{ 125 } \\ =\frac { 252 }{ 125 }

252 125 × 1000 = 2016 \frac { 252 }{ 125 } \times 1000=\boxed { 2016 }

The middle part was from one of solutions from my first question.

9 Helpful 0 Interesting 1 Brilliant 0 Confused

I was surprised that the answer was 2016.

BTW, you can simplify the sum as

∑ n = 0 ∞ ( n + 1 ) 2 6 n = 6 ∑ n = 1 ∞ n 2 6 n \sum _{n=0}^{\infty}\frac{\left(n+1\right)^2}{6^n}=6\sum _{n=1}^{\infty}\frac{n^2}{6^n}

Julian Poon - 5 years, 2 months ago

Log in to reply

I knew that!

So i wanted to add it in but i think i forgot.

Joel Yip - 5 years, 2 months ago

Wonderful problem in which 2016 comes again!

Soumava Pal - 5 years, 3 months ago

Level 5? How?

To be honest I only have a few Lvl 5 questions so I am happy.

Joel Yip - 5 years, 2 months ago

When you differentiate x n x^n wouldn't it be n . x n − 1 n. x^{n-1} ?

Anik Mandal - 5 years, 2 months ago

Log in to reply

Yes. This was from an earlier question. We took the derivative then multiplied by x x

Joel Yip - 5 years, 2 months ago

Differentiation=not required

Kaustubh Miglani - 5 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...