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Question: For non-negative integers s and r, let $\binom{s}{r} = \begin{cases} \frac{s!}{r!(s-r)!} & \text{if ...

For non-negative integers s and r, let (sr)={s!r!(s−r)!if r≤s0if r>s\binom{s}{r} = \begin{cases} \frac{s!}{r!(s-r)!} & \text{if } r \leq s \\ 0 & \text{if } r > s \end{cases}

For non-negative integers n and k, let Sn=∑k=0n(nk)2k+1k+1S_n = \sum_{k=0}^{n} \binom{n}{k} \frac{2^{k+1}}{k+1}

If e denotes the base of the natural logarithm, then which of the following is/are TRUE?

A

∑n=04(n+1)Sn=362\sum_{n=0}^{4} (n+1)S_n = 362

B

∑n=05(n+1)Sn=1086\sum_{n=0}^{5} (n+1)S_n = 1086

C

∑n=0∞Snn!=e3−e\sum_{n=0}^{\infty} \frac{S_n}{n!} = e^3 - e

D

∑n=0∞Snn!=e2−1\sum_{n=0}^{\infty} \frac{S_n}{n!} = e^2 - 1

Answer

(B), (C)

Explanation

Solution

  1. Simplify SnS_n: Sn=∑k=0n(nk)2k+1k+1S_n = \sum_{k=0}^{n} \binom{n}{k} \frac{2^{k+1}}{k+1} Using 1k+1(nk)=1n+1(n+1k+1)\frac{1}{k+1}\binom{n}{k} = \frac{1}{n+1}\binom{n+1}{k+1}: Sn=∑k=0n1n+1(n+1k+1)2k+1=1n+1∑k=0n(n+1k+1)2k+1S_n = \sum_{k=0}^{n} \frac{1}{n+1}\binom{n+1}{k+1} 2^{k+1} = \frac{1}{n+1} \sum_{k=0}^{n} \binom{n+1}{k+1} 2^{k+1} Let j=k+1j = k+1: Sn=1n+1∑j=1n+1(n+1j)2jS_n = \frac{1}{n+1} \sum_{j=1}^{n+1} \binom{n+1}{j} 2^{j} From the binomial expansion (1+2)n+1=∑j=0n+1(n+1j)2j=3n+1(1+2)^{n+1} = \sum_{j=0}^{n+1} \binom{n+1}{j} 2^j = 3^{n+1}: 3n+1=(n+10)20+∑j=1n+1(n+1j)2j=1+∑j=1n+1(n+1j)2j3^{n+1} = \binom{n+1}{0} 2^0 + \sum_{j=1}^{n+1} \binom{n+1}{j} 2^j = 1 + \sum_{j=1}^{n+1} \binom{n+1}{j} 2^j So, ∑j=1n+1(n+1j)2j=3n+1−1\sum_{j=1}^{n+1} \binom{n+1}{j} 2^j = 3^{n+1} - 1. Therefore, Sn=1n+1(3n+1−1)S_n = \frac{1}{n+1} (3^{n+1} - 1).

  2. Evaluate statements (A) and (B): (n+1)Sn=3n+1−1(n+1)S_n = 3^{n+1}-1. (A) ∑n=04(n+1)Sn=∑n=04(3n+1−1)=(31−1)+(32−1)+(33−1)+(34−1)+(35−1)\sum_{n=0}^{4} (n+1)S_n = \sum_{n=0}^{4} (3^{n+1}-1) = (3^1-1) + (3^2-1) + (3^3-1) + (3^4-1) + (3^5-1) =(3+9+27+81+243)−5=363−5=358= (3+9+27+81+243) - 5 = 363 - 5 = 358. Statement (A) is FALSE.

(B) ∑n=05(n+1)Sn=∑n=04(n+1)Sn+(5+1)S5=358+(35+1−1)=358+(36−1)=358+(729−1)=358+728=1086\sum_{n=0}^{5} (n+1)S_n = \sum_{n=0}^{4} (n+1)S_n + (5+1)S_5 = 358 + (3^{5+1}-1) = 358 + (3^6-1) = 358 + (729-1) = 358 + 728 = 1086. Statement (B) is TRUE.

  1. Evaluate statements (C) and (D): Snn!=3n+1−1(n+1)!\frac{S_n}{n!} = \frac{3^{n+1}-1}{(n+1)!}. ∑n=0∞Snn!=∑n=0∞3n+1−1(n+1)!\sum_{n=0}^{\infty} \frac{S_n}{n!} = \sum_{n=0}^{\infty} \frac{3^{n+1}-1}{(n+1)!}. Let m=n+1m = n+1. The sum becomes ∑m=1∞3m−1m!=∑m=1∞3mm!−∑m=1∞1m!\sum_{m=1}^{\infty} \frac{3^m-1}{m!} = \sum_{m=1}^{\infty} \frac{3^m}{m!} - \sum_{m=1}^{\infty} \frac{1}{m!}. Using the Maclaurin series for ex=∑m=0∞xmm!=1+∑m=1∞xmm!e^x = \sum_{m=0}^{\infty} \frac{x^m}{m!} = 1 + \sum_{m=1}^{\infty} \frac{x^m}{m!}, so ∑m=1∞xmm!=ex−1\sum_{m=1}^{\infty} \frac{x^m}{m!} = e^x - 1. ∑m=1∞3mm!=e3−1\sum_{m=1}^{\infty} \frac{3^m}{m!} = e^3 - 1. ∑m=1∞1m!=e1−1=e−1\sum_{m=1}^{\infty} \frac{1}{m!} = e^1 - 1 = e - 1. The total sum is (e3−1)−(e−1)=e3−e(e^3 - 1) - (e - 1) = e^3 - e.

(C) ∑n=0∞Snn!=e3−e\sum_{n=0}^{\infty} \frac{S_n}{n!} = e^3 - e. Statement (C) is TRUE. (D) ∑n=0∞Snn!=e2−1\sum_{n=0}^{\infty} \frac{S_n}{n!} = e^2 - 1. Statement (D) is FALSE because e3−e≠e2−1e^3-e \neq e^2-1.

The TRUE statements are (B) and (C).