Algebra Or Discrete Or Calculus?

Probability Level 4

L = lim ⁡ n → ∞ ∑ r = 1 n ∑ t = 0 r − 1 1 5 n ( n r ) ( r t ) 3 t L = \lim_{n\to \infty} \sum_{r=1}^{n} \sum_{t=0}^{r-1} \frac 1{5^n} \binom nr \binom rt 3^t

Given the above, find the value of L 99 + 9 9 L L^{99} + 99^L .

Notation: ( N M ) = N ! M ! ( N − M ) ! \dbinom NM = \dfrac {N!}{M!(N-M)!} denotes the binomial coefficient .


The answer is 100.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Chew-Seong Cheong
Apr 29, 2018

Relevant wiki: Binomial Theorem - Expansions - Basic

L = lim ⁡ n → ∞ ∑ r = 1 n ∑ t = 0 r − 1 1 5 n ( n r ) ( r t ) 3 t = lim ⁡ n → ∞ 1 5 n ∑ r = 1 n ( n r ) ∑ t = 0 r − 1 ( r t ) 3 t = lim ⁡ n → ∞ 1 5 n ∑ r = 1 n ( n r ) ( ∑ t = 0 r ( r t ) 3 t − 3 r ) By binomial theorem = lim ⁡ n → ∞ 1 5 n ∑ r = 1 n ( n r ) ( ( 1 + 3 ) r − 3 r ) = lim ⁡ n → ∞ 1 5 n ( ∑ r = 0 n ( n r ) ( 4 r − 3 r ) − ( n 0 ) ( 4 0 − 3 0 ) ) = lim ⁡ n → ∞ 1 5 n ∑ r = 0 n ( n r ) ( 4 r − 3 r ) = lim ⁡ n → ∞ 1 5 n ( ∑ r = 0 n ( n r ) 4 r − ∑ r = 0 n ( n r ) 3 r ) = lim ⁡ n → ∞ 1 5 n ( 5 n − 4 n ) = lim ⁡ n → ∞ 1 − ( 4 5 ) n = 1 \begin{aligned} L & = \lim_{n \to \infty} \sum_{r=1}^n \sum_{t=0}^{r-1} \frac 1{5^n}\binom nr \binom rt3^t \\ & = \lim_{n \to \infty} \frac 1{5^n} \sum_{r=1}^n \binom nr \sum_{t=0}^{\color{#3D99F6}r-1} \binom rt3^t \\ & = \lim_{n \to \infty} \frac 1{5^n} \sum_{r=1}^n \binom nr \left({\color{#3D99F6}\sum_{t=0}^{\color{#D61F06}r} \binom rt3^t} - 3^r\right) & \small \color{#3D99F6} \text{By binomial theorem} \\ & = \lim_{n \to \infty} \frac 1{5^n} \sum_{\color{#D61F06}r=1}^n \binom nr \left({\color{#3D99F6}(1+3)^r} - 3^r\right) \\ & = \lim_{n \to \infty} \frac 1{5^n} \left( \sum_{\color{#3D99F6}r=0}^n \binom nr \left(4^r - 3^r\right) - \binom n0 \left(4^0 - 3^0\right) \right) \\ & = \lim_{n \to \infty} \frac 1{5^n} \sum_{r=0}^n \binom nr \left(4^r - 3^r\right) \\ & = \lim_{n \to \infty} \frac 1{5^n} \left(\sum_{r=0}^n \binom nr 4^r - \sum_{r=0}^n \binom nr 3^r \right) \\ & = \lim_{n \to \infty} \frac 1{5^n} \left(5^n - 4^n \right) \\ & = \lim_{n \to \infty} 1- \left(\frac 45 \right)^n \\ & = 1 \end{aligned}

Therefore, L 99 + 9 9 L = 1 + 99 = 100 L^{99}+99^L = 1+99 = \boxed{100} .

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In the third line, when you change the upper limit of the summation from r − 1 r-1 to r r , shouldn't it say ∑ i = 0 r ( r t ) 3 t − 3 r \displaystyle\sum_{i=0}^r \binom{r}{t} 3^t - \color{#20A900} 3^r rather than ∑ i = 0 r ( r t ) 3 t − 1 \displaystyle\sum_{i=0}^r \binom{r}{t} 3^t - 1 ? The rest of the proof would still work with minor modifications.

zico quintina - 3 years, 1 month ago

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Thanks, you are right. I have changed the solution.

Chew-Seong Cheong - 3 years ago

The value of limit is equal to 1 i.e. L= 1 therefore L 99 + 9 9 L L^{99}+99^L = 100

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