Problems

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Solution:

\begin{align*} && V &= \lim_{n\rightarrow\infty}\left\{{1\over n+1} + {1\over n+2} + \cdots + {1\over n+n}\right\} \\ && &= \lim_{n\rightarrow\infty}\left\{\sum_{m=1}^n \frac{1}{n+m}\right\} \\ && &= \lim_{n\rightarrow\infty}\left\{\frac1n\sum_{m=1}^n \frac{1}{1+\frac{m}{n}}\right\} \\ &&&=\int_1^2 \frac{1}{x} \d x \\ &&&= \left [\ln x \right]_1^2 = \ln 2 \end{align*} \begin{align*} V &= \lim_{n\rightarrow\infty}\left\{{n\over n^2+1} + {n\over n^2+4} + \cdots + {n\over n^2+n^2}\right\} \\ &= \lim_{n\rightarrow\infty}\left\{\sum_{m=1}^n \frac{n}{n^2+m^2}\right\} \\ &= \lim_{n\rightarrow\infty}\left\{\frac{1}{n}\sum_{m=1}^n \frac{1}{1+\left (\frac{m}{n} \right)^2}\right\} \\ &= \int_0^1 \frac{1}{1+x^2} \d x \\ &= \left [\tan^{-1} x \right]_0^1 \\ &= \frac{\pi}4 \end{align*}

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Solution: Claim: \(u^2_{n+2} + u^2_{n-1} = 2( u^2_{n+1} + u^2_n )\) Proof: (By Induction). (Base Case): \(n = 1\) \begin{align*} && LHS &= u_{n+2}^2 + u_{n-1}^2 \\ &&&= u_3^2 + u_0^2 \\ &&&= 3^2 + 1^2 = 10\\ && RHS &= 2(u_{n+1}^2+u_n^2) \\ &&&= 2(2^2 + 1^2) \\ &&&= 10 \end{align*} Therefore the base case is true. (Inductive hypothesis) Suppose \(u^2_{n+2} + u^2_{n-1} = 2( u^2_{n+1} + u^2_n )\) is true for some \(n = k\), ie \(u^2_{k+2} + u^2_{k-1} = 2( u^2_{k+1} + u^2_k )\), the consider \(n = k+1\) \begin{align*} && LHS &= u_{k+1+2}^2 + u_{k+1-1}^2 \\ &&&= (u_{k+1}+u_{k+2})^2+u_k^2 \\ &&&= u_{k+2}^2+u_{k+1}^2+ u_k^2 + 2u_{k+1}u_{k+2} \\ &&&= u_{k+2}^2+u_{k+1}^2+ u_k^2 + 2u_{k+1}(u_{k+1}+u_k) \\ &&&= u_{k+2}^2 + u_{k+1}^2+u_k^2+2u_{k+1}^2+2u_{k+1}u_k \\ &&&= u_{k+1}^2+2u_{k+1}^2+ u_{k+1}^2+u_k^2+2u_{k+1}u_k \\ &&&= u_{k+2}^2+2u_{k+1}^2+ (u_{k+1}+u_k)^2 \\ &&&= u_{k+2}^2+2u_{k+1}^2+ u_{k+2}^2 \\ &&&=2(u_{k+2}^2+u_{k+1}^2) \\ &&&= RHS \end{align*} Therefore it is true for \(n = k+1\). Therefore by the principle of mathematical induction it is true for all \(n \geq 1\) Claim: $ u_{2n} = u^2_n + u^2_{n-1} \quad \mbox{ and }\quad u_{2n+1} = u^2_{n+1} - u^2_{n-1} $ Proof: Notice that \(\begin{pmatrix}u_{n+1} \\ u_n \end{pmatrix}= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n \begin{pmatrix}1 \\1 \end{pmatrix}\), in particular \begin{align*} && \begin{pmatrix}u_{n}& u_{n-1} \\ u_{n-1} & u_{n-2} \end{pmatrix}&= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n} \\ \Rightarrow && \begin{pmatrix}u_{2n}& u_{2n-1} \\ u_{2n-1} & u_{2n-2} \end{pmatrix}&= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{2n} \\ &&&= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n} \\ &&&=\begin{pmatrix}u_{n}& u_{n-1} \\ u_{n-1} & u_{n-2} \end{pmatrix}\begin{pmatrix}u_{n}& u_{n-1} \\ u_{n-1} & u_{n-2} \end{pmatrix}\\ &&&= \begin{pmatrix}u_{n}^2+u_{n-1}^2& u_{n-1}(u_n+u_{n-2}) \\ u_{n-1}(u_n+u_{n-2}) & u_{n-1}^2+u_{n-2}^2 \end{pmatrix} \end{align*} Therefore \(u_{2n} = u_{n}^2+u_{n-1}^2\) and \(u_{2n+1} = u_n(u_{n+1}+u_{n-1}) =(u_{n+1}-u_{n-1})(u_{n+1}-u_{n-1}) = u_{n+1}^2-u_{n-1}^2\)

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Solution: \begin{array}{c|c|c} n & a_n & b_n \\ \hline 1 & \cos x & 1 \\ \hline 2 & \frac12(1 + \cos x) & \sqrt{a_2} \\ &=\frac12(1+2\cos^2 \frac{x}{2}-1)& \sqrt{a_2} \\ &= \cos^2 \frac{x}{2} & \cos \frac{x}{2} \\ \hline 3 & \frac12(\cos^2 \frac{x}{2}+\cos \frac{x}{2}) & \sqrt{a_3\cos \frac{x}{2}} \\ &= \cos \frac{x}{2} \cdot \frac12 (\cos \frac{x}{2}+1) & \sqrt{a_3\cos \frac{x}{2}} \\ &= \cos \frac{x}{2} \cos^2 \frac{x}{4} & \cos \frac{x}{2} \cos \frac{x}{4} \end{array} Claim: \(\displaystyle a_n = \cos \frac{x}{2^{n-1}}\prod_{k=1}^{n-1} \cos \frac{x}{2^k}\), \(\displaystyle b_n = \prod_{k=1}^{n-1} \cos \frac{x}{2^k}\) Claim: \(a_{n+1} = \frac12(a_n + b_n)\) Proof: \begin{align*} && \frac12(a_n + b_n) &= \frac12 \left ( \cos \frac{x}{2^{n-1}}\prod_{k=1}^{n-1} \cos \frac{x}{2^k} + \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \right) \\ &&&= \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \frac12\left (\cos \frac{x}{2^{n-1}} + 1 \right) \\ &&&= \left ( \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \right) \cos^{2} \frac{x}{2^n} \\ &&&= \cos \frac{x}{2^n} \prod_{k=1}^{n} \cos \frac{x}{2^k} \\ &= a_{n+1} \end{align*} Claim: \(b_{n+1} = \sqrt{a_{n+1}b_n}\) Proof: \begin{align*} && \sqrt{a_{n+1}b_n} &= \sqrt{ \cos \frac{x}{2^n} \prod_{k=1}^{n} \cos \frac{x}{2^k} \cdot \prod_{k=1}^{n-1} \cos \frac{x}{2^k} }\\ &&&= \prod_{k=1}^{n-1} \cos \frac{x}{2^k} \sqrt{\cos ^2\frac{x}{2^{n}}} \\ &&&= \prod_{k=1}^{n} \cos \frac{x}{2^k} \\ &&&= b_{n+1} \end{align*}

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Solution: First notice by partial fractions: \begin{align*} \frac{2n-1}{n(n+1)(n+2)} &= \frac{-1/2}{n} + \frac{3}{n+1} + \frac{-5/2}{n+2} \\ &= \frac{-1}{2n} + \frac{3}{n+1} - \frac{5}{2(n+2)} \end{align*} And therefore: \begin{align*} \sum_{n = 1}^N \frac{2n-1}{n(n+1)(n+2)} &= -\frac12 \sum_{n=1}^N \frac1n +3\sum_{n=1}^N \frac1{n+1} -\frac52 \sum_{n=1}^N \frac1{n+2} \\ &= -\frac12-\frac14 + \frac{3}{2}+ \sum_{n=3}^N (3-\frac12 -\frac52)\frac1n + \frac{3}{N+1} - \frac{5}{2(N+1)} - \frac{5}{2(N+2)} \\ &= \frac12 \l \frac32+\frac1{N+1}-\frac{5}{N+2} \r \end{align*} As \(N \to \infty, S_N \to \frac{3}{4}\). \begin{align*} && \frac{a_n}{a_{n-1}}&=\frac{(n-1)(2n-1)}{(n+2)(2n-3)} \\ \Rightarrow && \frac{a_n}{a_1} &= \frac{a_n}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_2}{a_1} \\ &&&= \frac{(n-1)(2n-1)}{(n+2)(2n-3)} \cdot \frac{(n-2)(2n-3)}{(n+1)(2n-5)} \cdots \frac{(1)(3)}{(4)(1)} \\ &&&= \frac{(2n-1)3\cdot 2\cdot 1}{(n+2)(n+1)n} \\ &&& = \frac{6(2n-1)}{n(n+1)(n+2)} \end{align*} Therefore \(a_n = \frac{4}{3} \frac{2n-1}{n(n+1)(n+2)}\) and so our sequence is \(\frac43\) the earlier sum, ie \(1\)

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Solution: Suppose \(V_T = Ak^T + B\), then \begin{align*} && Ak^{T+1}+B &= (1+c)(Ak^T+B) - d \\ \Rightarrow && (k-1-c)Ak^T &= cB -d \\ \Rightarrow && k &= 1+c \\ && B &= \frac{d}{c} \\ && A &= V_0 - B \\ \Rightarrow && V_N &= (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c} \end{align*} When \(c = 0\), \(V_{N+1} = V_N - d \Rightarrow V_N = V_0 - Nd\). \begin{align*} \lim_{c \to 0} \left ( (V_0 - \frac{d}{c})(1+c)^{N} + \frac{d}{c} \right) &= \lim_{c \to 0} \left ( \frac{(V_0 c-d)(1+c)^N + d}{c} \right ) \\ &= \lim_{c \to 0} \left ( \frac{V_0c - d-Ncd+NV_0c^2 + o(c^2) + d}{c} \right ) \\ &= \lim_{c \to 0} \left ( V_0 - Nd + o(c) \right ) \\ &= V_0 - Nd \end{align*}

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Solution:

1. Notice that \(a_n = 3^{3^{n-1}}\) in particular, \(a_7 = 3^{3^6}\). Using Fermat's little theorem, we can see that \(3^4 \equiv 1 \pmod{5}\) and so we need to figure out \(3^6 \pmod{4}\), which is clearly \(1\). Therefore \(3^{3^6} \equiv 3^{4k+1} \equiv 3 \pmod{5}\). Therefore the units digit is \(3\).
2. Notice that \(3^5 > 100\) and \(3^3 > 10\). Therefore \begin{align*} a_7 &= 3^{3^6} \\ &= (3^3)^{3^5} \\ &> 10^{3^5} \\ &> 10^{100} \end{align*}
3. \begin{align*} \frac{a_7+1}{2a_7} &= \frac12 + \frac1{2a_7} \\ &= 0.5 + 0.\underbrace{000\cdots}_{\text{at least }99\text{ zeros}} \\ &= 0.50 \end{align*} Since \(a_7 > 10^{100}, \, \frac{1}{2a_7} < 10^{-100}\)

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Solution: The first digit is \(2\) or \(5\), all the other digits can be any value from \(0,2,5,7\). Therefore we have \begin{align*} S &= 2000 \cdot 4^3+5000 \cdot 4^3 + (200+500+700) \cdot 2 \cdot 4^2 + (20+50+70) \cdot 2 \cdot 4^2 + (2+5+7) \cdot 2 \cdot 4^2 \\ &= 7 \cdot 4^3 \cdot 2^3 \cdot 5^3 + 14 \cdot 2 \cdot 4^2 \cdot 111 \\ &= 2^{9} \cdot5^3 \cdot 7 + 2^{6} \cdot 3 \cdot 7 \cdot 37 \\ &= 2^6 \cdot 7 \cdot (1000+111) \\ &= 2^6 \cdot 7 \cdot 11 \cdot 101 \end{align*} Alternatively, consider adding the first and last terms, and second and second and last terms, etc we obtain \(7777\). There are \(2 \cdot 4^3\) terms so \(7777 \cdot 4^3 = 2^6 \cdot 7 \cdot 11 \cdot 101\)

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Solution: \begin{align*} x_1 &= \frac{2}{3} \\ x_n &= \frac{2}{3-x_{n-1}} \\ x_2 &= \frac{2}{3 - \frac23} \\ &= \frac{6}7 \\ x_3 &= \frac{2}{3-\frac67} \\ &= \frac{14}{15} \\ x_4 &= \frac{2}{3 - \frac{14}{15}} \\ &= \frac{30}{31} \end{align*} Guess: \(x_n = \frac{2^{n+1}-2}{2^{n+1}-1}\). Proof: (By induction) (Base case): We have checked several initial cases. (Inductive step): Suppose our formula is true for \(n = k\), then consider: \begin{align*} x_{k+1} &= \frac{2}{3 - x_{k}} \\ &= \frac{2}{3 - \frac{2^{k+1}-2}{2^{k+1}-1}}\tag{assumption} \\ &= \frac{2\cdot(2^{k+1}-1)}{3 \cdot(2^{k+1}-1) - (2^{k+1}-2) } \\ &= \frac{2^{k+2}-2}{2\cdot 2^{k+1} - 3 + 2 } \\ &= \frac{2^{k+2}-2}{ 2^{k+2} - 1 } \\ \end{align*} Therefore, if our formula is true for \(n = k\) it is true for \(n = k+1\). Therefore by the principle of mathematical induction it is true for \(n \geq 1, n \in \mathbb{Z}\)

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Solution:

1. \(\,\) \begin{align*} && \sum_{r=1}^n \left (f(r) - f(r-1) \right) &= \cancel{f(1)} - f(0) + \cdots\\ &&&\quad\, \cancel{\f(2)}-\cancel{f(1)} + \cdots \\ &&&\quad\, \cancel{f(3)}-\cancel{f(2)} + \cdots \\ &&&\quad\, +\cdots + \cdots \\ &&&\quad\, \cancel{f(n-1)}-\cancel{f(n-2)} + \cdots \\ &&&\quad\, f(n)-\cancel{f(n-1)} \\ &&&=f(n) - f(0) \end{align*}
2. If \(f(r) = r^2(r+1)^2\) then \begin{align*} && f(r) - f(r-1) &= r^2(r+1)^2 - (r-1)r^2 \\ &&&= r^2((r+1)^2-(r-1)^2) \\ &&&=4r^3 \end{align*} Therefore \begin{align*} && \sum_{r=1}^n 4r^3 &= n^2(n+1)^2-0 \\ \Rightarrow && \sum_{r=1}^n r^3 &= \frac{n^2(n+1)^2}{4} \\ \end{align*}
3. So \begin{align*} && \sum_{r=1}^{2n+1} (-1)^{r-1}r^3 &= \sum_{r=1}^{2n+1} r^3 - 2 \sum_{r=1}^n (2r)^3 \\ &&&= \frac14(2n+1)^2(2n+2)^2 - 8 \frac{n^2(n+1)^2}{4} \\ &&&= (2n+1)^2(n+1)^2 - 2n^2(n+1)^2 \\ &&&= (n+1)^2(4n^2+4n+1 - 2n^2) \\ &&&= (n+1)^2(2n^2+4n+1) \end{align*}

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Solution: \begin{align*} && (n^2+1) + (n^2+2) + \cdots + (n+1)^2 &= n^3+(n+1)^3 \\ \Leftrightarrow && \sum_{i=n^2+1}^{(n+1)^2} i &= n^3 + (n+1)^3 \\ && \sum_{i=n^2+1}^{(n+1)^2} i &= \sum_{i=1}^{(n+1)^2} i- \sum_{i=1}^{n^2} i \\ &&&= \frac{(n+1)^2((n+1)^2+1)}{2} - \frac{n^2(n^2+1)}{2} \\ &&&= \frac{(n+1)^2(n^2+2n+2) - n^2(n^2+1)}{2} \\ &&&= \frac{2(n+1)^3+n^2(n^2+2n+1) - n^2(n^2+1)}{2}\\ &&&= \frac{2(n+1)^3+2n^3 + n^2(n^2+1) - n^2(n^2+1)}{2}\\ &&&= (n+1)^3+n^3 \end{align*} \begin{align*} && \sum_{i=1}^{N^2} i &=(0^3+1^3)+ (1^3+2^3)+(2^3+3^3) + \cdots + ((N-1)^3+N^3) \\ &&&= 2 \left (1^3+2^3 + 3^3 + \cdots + (N-1)^3 \right) + N^3 \\ \Rightarrow && \sum_{i=1}^N i^3 &= \frac12 \left ( N^3+ \sum_{i=1}^{N^2} i \right) \\ &&&= \frac12 \left ( N^3 + \frac{N^2(N^2+1)}{2} \right) \\ &&&= \frac{N^2(N^2+1)+2N^3}{4} \\ &&&= \frac{N^2(N^2+2N+1)}{4} \\ &&&= \frac{N^2(N+1)^2}{4} \\ \end{align*}

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Solution: \begin{align*} && 0 &\leqslant (a_1 - a_2)^2 \\ &&&= a_1^2 -2a_1a_2 + a_2^2 \\ &&&= (a_1+a_2)^2 -4a_1a_2 \\ \Leftrightarrow && a_1a_2 &\leqslant \left ( \frac{a_1+a_2}2 \right)^2 \end{align*} Claim: \((*)\) is true Proof: (By induction) We have already proven the base case. Suppose it is true for some \(m\), then consider \(m+1\) \begin{align*} && a_1 \cdots a_{2^m} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^m}}{2^m} \right)^{2^m} \tag{by (*)} \\ && a_{2^m+1} \cdots a_{2^{m+1}} &\leqslant \left ( \frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} \right)^{2^m} \tag{by (*)} \\ \Rightarrow && (a_1 \cdots a_{2^m})^{1/2^m} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^m}}{2^m} \right) \\ && (a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} &\leqslant \left ( \frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} \right) \\ \Rightarrow && (a_1 \cdots a_{2^m})^{1/2^m} \cdot (a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} &\leqslant \left ( \frac{ (a_1 \cdots a_{2^m})^{1/2^m} +(a_{2^m+1} \cdots a_{2^{m+1}})^{1/2^m} }{2} \right )^2 \\ &&&\leqslant \left ( \frac{ \frac{a_1 + \cdots + a_{2^m}}{2^m}+\frac{a_{2^m+1} + \cdots + a_{2^{m+1}}}{2^m} }{2} \right )^2 \\ &&&\leqslant \left ( \frac{ a_1 + \cdots + a_{2^m}+a_{2^m+1} + \cdots + a_{2^{m+1}} }{2^{m+1}} \right )^2 \\ \Rightarrow && a_1 \cdots a_{2^{m+1}} &\leqslant \left ( \frac{a_1 + \cdots + a_{2^{m+1}}}{2^{m+1}} \right)^{2^{m+1}} \end{align*} Which is precisely \((*)\) for \(m+1\). Therefore our statement is true by induction. Suppose \(n < 2^m\) and \(b_1 = a_1, b_2 = a_2, \cdots b_n = a_n\) and \(b_{n+1} = \cdots = b_{2^m} = A\) where \(A = \frac{a_1 + \cdots + a_n}{n}\) then \begin{align*} && b_1 \cdots b_n \cdot b_{n+1} \cdots b_{2^m} &\leq \left ( \frac{b_1 + \cdots + b_n + b_{n+1} + \cdots + b_{2^m}}{2^{m}} \right)^{2^m} \\ \Leftrightarrow && a_1 \cdots a_n \cdot A^{2^m-n} &\leq \left ( \frac{a_1 + \cdots + a_n + (2^m-n)A}{2^m} \right)^{2^m} \\ &&&= \left ( \frac{nA + (2^m - n)A}{2^m} \right)^{2^m} \\ &&&= A^{2^m} \\ \Rightarrow && a_1 \cdots a_n &\leq A^n \\ \Rightarrow && (a_1 \cdots a_n)^{1/n} &\leq A = \frac{a_1 + \cdots + a_n}{n} \end{align*}