Color wash the series

Calculus Level 3

1 2 1 + 5 2 11 + 9 2 ( 11 ) 2 + 1 3 2 ( 11 ) 3 + 1 7 2 ( 11 ) 4 + ⋯ = ? \dfrac{\color{magenta}{1^2}}{\color{#302B94}{1}}+\dfrac{\color{magenta}{5^2}}{\color{#302B94}{11}}+\dfrac{\color{magenta}{9^2}}{\color{#302B94}{(11)^2}}+\dfrac{\color{magenta}{13^2}}{\color{#302B94}{(11)^3}}+\dfrac{\color{magenta}{17^2}}{\color{#302B94}{(11)^4}}+\cdots= \, ?


The answer is 4.092.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Ikkyu San
Aug 9, 2015

Let

S 1 = 1 2 1 + 5 2 11 + 9 2 1 1 2 + 1 3 2 1 1 3 + 1 7 2 1 1 4 + ⋯ ⇒ ( 1 ) \begin{aligned}\color{magenta}{S_1=}&\ \color{magenta}{\dfrac{1^2}1+\dfrac{5^2}{11}+\dfrac{9^2}{11^2}+\dfrac{13^2}{11^3}+\dfrac{17^2}{11^4}+\cdots}\Rightarrow(1)\end{aligned}

Multiplying with 1 11 \dfrac1{11} in equation ( 1 ) (1)

1 11 S 1 = 1 2 11 + 5 2 1 1 2 + 9 2 1 1 3 + 1 3 2 1 1 4 + 1 7 2 1 1 5 + ⋯ ⇒ ( 2 ) \begin{aligned}\dfrac1{11}\color{magenta}{S_1}=&\ \dfrac{1^2}{11}+\dfrac{5^2}{11^2}+\dfrac{9^2}{11^3}+\dfrac{13^2}{11^4}+\dfrac{17^2}{11^5}+\cdots\Rightarrow(2)\end{aligned}

Subtracting equation ( 1 ) (1) with equation ( 2 ) (2)

( 1 − 1 11 ) S 1 = 1 2 1 + 5 2 − 1 2 11 + 9 2 − 5 2 1 1 2 + 1 3 2 − 9 2 1 1 3 + 1 7 2 − 1 3 2 1 1 4 + ⋯ 10 11 S 1 = 1 + 4 × 6 11 + 4 × 14 1 1 2 + 4 × 22 1 1 3 + 4 × 30 1 1 4 + ⋯ 10 11 S 1 = 1 + 4 ( 6 11 + 14 1 1 2 + 22 1 1 3 + 30 1 1 4 + ⋯ ) ⇒ ( 3 ) \begin{aligned}\left(1-\dfrac1{11}\right)\color{magenta}{S_1}=&\ \dfrac{1^2}{1}+\dfrac{5^2-1^2}{11}+\dfrac{9^2-5^2}{11^2}+\dfrac{13^2-9^2}{11^3}+\dfrac{17^2-13^2}{11^4}+\cdots\\\dfrac{10}{11}\color{magenta}{S_1}=&\ 1+\dfrac{\color{teal}4\times6}{11}+\dfrac{\color{teal}4\times14}{11^2}+\dfrac{\color{teal}4\times22}{11^3}+\dfrac{\color{teal}4\times30}{11^4}+\cdots\\\dfrac{10}{11}\color{magenta}{S_1}=&\ 1+\color{teal}4\color{#302B94}{\left(\dfrac6{11}+\dfrac{14}{11^2}+\dfrac{22}{11^3}+\dfrac{30}{11^4}+\cdots\right)}\Rightarrow(3)\end{aligned}

Let

S 2 = 6 11 + 14 1 1 2 + 22 1 1 3 + 30 1 1 4 + ⋯ ⇒ ( 4 ) \begin{aligned}\color{#302B94}{S_2=}&\ \color{#302B94}{\dfrac6{11}+\dfrac{14}{11^2}+\dfrac{22}{11^3}+\dfrac{30}{11^4}+\cdots}\Rightarrow(4)\end{aligned}

Multiplying with 1 11 \dfrac1{11} in equation ( 4 ) (4)

1 11 S 2 = 6 1 1 2 + 14 1 1 3 + 22 1 1 4 + 30 1 1 5 + ⋯ ⇒ ( 5 ) \begin{aligned}\dfrac1{11}\color{#302B94}{S_2}=&\ \dfrac6{11^2}+\dfrac{14}{11^3}+\dfrac{22}{11^4}+\dfrac{30}{11^5}+\cdots\Rightarrow(5)\end{aligned}

Subtracting equation ( 4 ) (4) with equation ( 5 ) (5)

( 1 − 1 11 ) S 2 = 6 11 + 14 − 6 1 1 2 + 22 − 14 1 1 3 + 30 − 22 1 1 4 + ⋯ 10 11 S 2 = 6 11 + ( 8 1 1 2 + 8 1 1 3 + 8 1 1 4 + ⋯ ) 10 11 S 2 = 6 11 + ( 8 1 1 2 1 − 1 11 ) 10 11 S 2 = 6 11 + ( 8 121 × 11 10 ) 10 11 S 2 = 6 11 + 4 55 S 2 = 30 + 4 55 × 11 10 = 17 25 \begin{aligned}\left(1-\dfrac1{11}\right)\color{#302B94}{S_2}=&\ \dfrac6{11}+\dfrac{14-6}{11^2}+\dfrac{22-14}{11^3}+\dfrac{30-22}{11^4}+\cdots\\\dfrac{10}{11}\color{#302B94}{S_2}=&\ \dfrac6{11}+\left(\dfrac8{11^2}+\dfrac8{11^3}+\dfrac8{11^4}+\cdots\right)\\\dfrac{10}{11}\color{#302B94}{S_2}=&\ \dfrac6{11}+\left(\dfrac{\frac8{11^2}}{1-\frac1{11}}\right)\\\dfrac{10}{11}\color{#302B94}{S_2}=&\ \dfrac6{11}+\left(\dfrac8{121}\times\dfrac{11}{10}\right)\\\dfrac{10}{11}\color{#302B94}{S_2}=&\ \dfrac6{11}+\dfrac4{55}\\\color{#302B94}{S_2}=&\ \dfrac{30+4}{55}\times\dfrac{11}{10}=\color{#302B94}{\dfrac{17}{25}}\end{aligned}

Instead S 2 \color{#302B94}{S_2} with 17 25 \color{#302B94}{\dfrac{17}{25}} in equation ( 3 ) (3)

10 11 S 1 = 1 + 4 ( 17 25 ) 10 11 S 1 = 1 + 68 25 S 1 = 25 + 68 25 × 11 10 = 1023 250 = 4.092 \begin{aligned}\dfrac{10}{11}\color{magenta}{S_1}=&\ 1+\color{teal}4\color{#302B94}{\left(\dfrac{17}{25}\right)}\\\dfrac{10}{11}\color{magenta}{S_1}=&\ 1+\dfrac{68}{25}\\\color{magenta}{S_1}=&\ \dfrac{25+68}{25}\times\dfrac{11}{10}=\color{magenta}{\dfrac{1023}{250}}=\boxed{4.092}\end{aligned}

21 Helpful 2 Interesting 0 Brilliant 0 Confused

Moderator note:

Very neat work as always. Wonderful!

This is an Arithmetic-Geometric Progression approach whereby we extend the summation method that takes advantage of Method of Differences.

For the sake of variety, and for a simpler solution. Can you think of a simpler approach?

Hint : Calculus. Differentiate ∑ n = 0 ∞ x n = 1 1 − x \displaystyle \sum_{n=0}^\infty x^n = \frac1{1-x} .

As challenge master said,

The CALCULUS METHOD

The required summation is,

∑ 0 ∞ ( 4 n + 1 ) 2 1 1 n = 16 ∑ 0 n 2 1 1 n + 8 ∑ 0 n 1 1 n + ∑ 0 1 1 1 n \displaystyle \sum_0^\infty \frac{(4n +1)^2}{11^{n}} = 16\sum_0\frac{n^2}{11^n} + 8\sum_0\frac{n}{11^n} + \sum_0 \frac{1}{11^n}

Now,

Let A ( x ) = 1 + x + x 2 + x 3 + ⋯ = ( 1 − x ) − 1 A(x) = 1 + x + x^2 + x^3 + \cdots = (1-x)^{-1}

⇒ ∑ 0 a n ⋅ x n = ( 1 − x ) − 1 , a n = 1 \Rightarrow \displaystyle \sum_0 a_n\cdot x^n = (1-x)^{-1}, a_n = 1

Now we want

1 x + 2 x 2 + 3 x 3 + ⋯ = ∑ 0 n ⋅ x n = ∑ 0 b n ⋅ x n , b n = n a n \displaystyle 1x + 2x^2 + 3x^3 + \cdots = \sum_0 n\cdot x^n = \sum_0 b_n \cdot x^n, b_n = n a_n

1 x + 2 x 2 + 3 x 3 + ⋯ = x ⋅ d d x ( 1 + x + x 2 + x 3 + ⋯ ) = x ⋅ d d x ( 1 − x ) − 1 = x ( 1 − x ) − 2 \displaystyle \begin{aligned} 1x + 2x^2 + 3x^3 + \cdots &= x \cdot \frac{d}{dx}(1 + x + x^2 + x^3 + \cdots)\\ &=x\cdot \frac{d}{dx}(1-x)^{-1}\\ &=x(1-x)^{-2}\end{aligned}

⇒ ∑ 0 b n ⋅ x n = x ( 1 − x ) − 2 \displaystyle \Rightarrow \sum_0 b_n \cdot x^n = x(1-x)^{-2}

Then,

1 2 x + 2 2 x 2 + 3 2 x 3 + ⋯ = ∑ 0 n 2 ⋅ x n = ∑ 0 c n ⋅ x n , c n = n 2 a n \displaystyle 1^2x + 2^2x^2 + 3^2x^3 + \cdots = \sum_0 n^2\cdot x^n = \sum_0 c_n \cdot x^n, c_n = n^2a_n

1 2 x + 2 2 x 2 + 3 2 x 3 + ⋯ = x ⋅ d d x ( x + 2 x 2 + 3 x 3 + ⋯ ) = x ⋅ d d x x ( 1 − x ) − 2 = x [ 2 x ( 1 − x ) − 3 + ( 1 − x ) − 2 ] = x ( 1 − x ) − 2 [ 1 + x 1 − x ] \displaystyle \begin{aligned} 1^2x + 2^2x^2 + 3^2x^3 + \cdots &= x \cdot \frac{d}{dx}( x + 2x^2 + 3x^3 + \cdots)\\ &=x\cdot \frac{d}{dx}x(1-x)^{-2}\\ &=x\left[2x(1-x)^{-3} + (1-x)^{-2}\right]\\ &=x(1-x)^{-2}\left[\frac{1+x}{1-x}\right]\end{aligned}

⇒ ∑ 0 c n ⋅ x n = x ( 1 − x ) − 2 [ 1 + x 1 − x ] \displaystyle \Rightarrow \sum_0 c_n \cdot x^n = x(1-x)^{-2}\left[\frac{1+x}{1-x}\right]

Lets get back!

∑ 0 ∞ ( 4 n + 1 ) 2 1 1 n = 16 ∑ 0 n 2 1 1 n + 8 ∑ 0 n 1 1 n + ∑ 0 1 1 1 n = 16 ∑ 0 c n ⋅ x n + 8 ∑ 0 b n ⋅ x n + ∑ 0 b n ⋅ x n , x = 1 11 = 16 11 121 100 12 10 + 8 11 121 100 + 11 10 = 16 × 11 × 12 + 11 × 80 + 1100 1000 = 4.092 \displaystyle \begin{aligned}\sum_0^\infty \frac{(4n +1)^2}{11^{n}} &= 16\sum_0\frac{n^2}{11^n} + 8\sum_0\frac{n}{11^n} + \sum_0 \frac{1}{11^n}\\ &= 16\sum_0 c_n \cdot x^n + 8 \sum_0 b_n \cdot x^n + \sum_0 b_n \cdot x^n,&x = \frac{1}{11}\\ \\ &= \frac{16}{11}\frac{121}{100}\frac{12}{10} + \frac{8} {11}\frac{121}{100} + \frac{11}{10}\\ \\ &= \frac{16\times 11 \times 12 + 11\times 80 + 1100 }{1000}\\\\ &=\boxed{4.092}\end{aligned}

Kishore S. Shenoy - 5 years, 10 months ago

Love this!!

KATHLEEN KASPER - 5 years, 10 months ago

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