Power 3

Algebra Level 3

∑ n = 1 ∞ n 3 3 n = X \sum_{n=1}^ \infty \frac{ { n }^{ 3 }}{ 3^n } = X

X X can be expressed in the form a b \dfrac{a}{b} , where a a and b b are coprime positive integers. Find a + b a + b .

Too easy? Go to the next level


The answer is 41.

Saharsh Rathi by saharsh rathi , 22, India

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2 solutions

X = ∑ n = 1 ∞ n 3 3 n = ∑ n = 0 ∞ n 3 3 n = ∑ n = 0 ∞ ( n + 1 ) 3 3 n + 1 = ∑ n = 0 ∞ n 3 + 3 n 2 + 3 n + 1 3 n + 1 = ∑ n = 0 ∞ n 3 3 n + 1 + 3 ∑ n = 0 ∞ n 2 3 n + 1 + 3 ∑ n = 0 ∞ n 3 n + 1 + ∑ n = 0 ∞ 1 3 n + 1 = 1 3 ∑ n = 0 ∞ n 3 3 n + 3 3 ∑ n = 0 ∞ n 2 3 n + 3 3 ∑ n = 0 ∞ n 3 n + 1 3 ∑ n = 0 ∞ 1 3 n = X 3 + 3 2 + 3 4 + 1 3 ( 1 1 − 1 3 ) See Note. 2 3 X = 11 4 X = 33 8 \begin{aligned} X & =\sum_{\color{#3D99F6}n=1}^\infty \frac {n^3}{3 ^n} =\sum_{\color{#D61F06}n=0} ^\infty \frac {n^3}{3 ^n} \\ & =\sum_{\color{#D61F06}n=0} ^\infty \frac {({\color{#D61F06}n+1})^3}{3 ^{\color{#D61F06}n+1}} \\ & =\sum_{\color{#D61F06}n=0} ^\infty \frac {n^3+3n^2+3n+1}{3 ^{n+1}} \\ & = \sum_{\color{#D61F06}n=0} ^\infty \frac {n^3}{3^{n+1}} + 3 \sum_{\color{#D61F06}n=0} ^\infty \frac {n^2}{3 ^{n+1}} + 3 \sum_{\color{#D61F06}n=0}^\infty \frac n{3 ^{n+1}} + \sum_{\color{#D61F06}n=0}^\infty \frac 1{3 ^{n+1}} \\ & = \frac 13 \sum_{\color{#D61F06}n=0}^\infty \frac {n^3}{3^n} + \frac 33 \sum_{\color{#D61F06}n=0}^\infty \frac {n^2}{3 ^n} + \frac 33 \sum_{\color{#D61F06}n=0}^\infty \frac n{3 ^n} + \frac 13 \sum_{\color{#D61F06}n=0}^\infty \frac 1{3 ^n} \\ & = \frac X3 + {\color{#3D99F6}\frac 32} + {\color{#3D99F6}\frac 34} + \frac 13 \left(\frac 1 {1 - \frac 1 3} \right) & \small {\color{#3D99F6} \text{See Note.}} \\ \frac 23 X & = \frac {11}4 \\ X & = \frac {33}8 \end{aligned}

⟹ a + b = 33 + 8 = 41 \implies a+b =33+8=\boxed{41}

Note: Using similar technique:

X 1 = ∑ n = 0 ∞ n 3 n = ∑ n = 1 ∞ n 3 n = ∑ n = 0 ∞ n + 1 3 n + 1 = ∑ n = 0 ∞ n 3 n + 1 + ∑ n = 0 ∞ 1 3 n + 1 = 1 3 ∑ n = 0 ∞ n 3 n + 1 3 ∑ n = 0 ∞ 1 3 n = X 1 3 + 1 3 ( 1 1 − 1 3 ) 2 3 X 1 = 1 2 ⟹ X 1 = 3 4 \begin{aligned} X_1 & = \sum_{\color{#D61F06}n=0} ^\infty \frac n{3 ^n} \\ & = \sum_{\color{#3D99F6}n=1}^\infty \frac n{3 ^n} \\ & =\sum_{\color{#D61F06}n=0}^\infty \frac {\color{#D61F06}n+1}{3 ^{\color{#D61F06}n+1}} \\ & = \sum_{\color{#D61F06}n=0}^\infty \frac n{3 ^{n+1}} + \sum_{\color{#D61F06}n=0}^\infty \frac 1{3 ^{n+1}} \\ & = \frac 13 \sum_{\color{#D61F06}n=0}^\infty \frac n{3 ^n} + \frac 13 \sum_{\color{#D61F06}n=0}^\infty \frac 1{3 ^n} \\ & = \frac {X_1}3 + \frac 13\left(\frac 1 {1 - \frac 1 3} \right) \\ \frac 23 X_1 & = \frac 12 \\ \implies X_1 & = \frac 34 \end{aligned}

X 2 = ∑ n = 0 ∞ n 2 3 n = ∑ n = 1 ∞ n 2 3 n = ∑ n = 0 ∞ ( n + 1 ) 2 3 n + 1 = ∑ n = 0 ∞ n 2 + 2 n + 1 3 n + 1 = ∑ n = 0 ∞ n 2 3 n + 1 + 2 ∑ n = 0 ∞ n 3 n + 1 + ∑ n = 0 ∞ 1 3 n + 1 = 1 3 ∑ n = 0 ∞ n 2 3 n + 2 3 ∑ n = 0 ∞ n 3 n + 1 3 ∑ n = 0 ∞ 1 3 n = X 2 3 + 2 3 ⋅ 3 4 + 1 3 ( 1 1 − 1 3 ) X 1 = 3 4 2 3 X 2 = 1 X 2 = 3 2 \begin{aligned} X_2 & =\sum_{\color{#D61F06}n=0} ^\infty \frac {n^2}{3 ^n} \\ & =\sum_{\color{#3D99F6}n=1}^\infty \frac {n^2}{3 ^n} \\ & =\sum_{\color{#D61F06}n=0} ^\infty \frac {({\color{#D61F06}n+1})^2}{3 ^{\color{#D61F06}n+1}} \\ & =\sum_{\color{#D61F06}n=0} ^\infty \frac {n^2+2n+1}{3 ^{n+1}} \\ & = \sum_{\color{#D61F06}n=0} ^\infty \frac {n^2}{3 ^{n+1}} + 2 \sum_{\color{#D61F06}n=0}^\infty \frac n{3 ^{n+1}} + \sum_{\color{#D61F06}n=0}^\infty \frac 1{3 ^{n+1}} \\ & = \frac 13 \sum_{\color{#D61F06}n=0}^\infty \frac {n^2}{3 ^n} + \frac 23 \sum_{\color{#D61F06}n=0}^\infty \frac n{3 ^n} + \frac 13 \sum_{\color{#D61F06}n=0}^\infty \frac 1{3 ^n} \\ & = \frac {X_2}3 + \frac 23 \cdot {\color{#3D99F6}\frac 34} + \frac 13 \left(\frac 1 {1 - \frac 1 3} \right) & \small {\color{#3D99F6} X_1 = \frac 34} \\ \frac 23 X_2 & = 1 \\ X_2 & = \frac 32 \end{aligned}

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Woah ... So big...

Md Zuhair - 4 years, 7 months ago

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Where is level 5 ?

Sabhrant Sachan - 4 years, 7 months ago

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https://brilliant.org/problems/level-5-2/

saharsh rathi - 4 years, 7 months ago

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@Saharsh Rathi – Check out another method in Level 5.

Chew-Seong Cheong - 4 years, 7 months ago

@Chew-Seong Cheong BEST SOLUTION . This method makes this question very easy. Thanks for letting us know such a great method to solve this. I also was able to solve power 4 , which i couldn't solve with my way because it was getting too complicated. :)

saharsh rathi - 4 years, 7 months ago

  • syms n

  • symsum(n^3/(3^n), n, 0, Inf)

  • ans =

  • 33/8

Using Matlab :P

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