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Algebra Difficulty 5.7 AIME, harder Find the answer

10. (25 points) For any positive integers m,nm, n, define the function f(m,n)f(m, n) as follows:
(i) f(1,1)=1f(1,1)=1;
(ii) f(m+1,n)=f(m,n)+2(m+n)f(m+1, n)=f(m, n)+2(m+n);
(iii) f(m,n+1)=f(m,n)+2(m+n−1)f(m, n+1)=f(m, n)+2(m+n-1).
(1) Find the analytical expression for f(m,n)f(m, n);
(2) Let an=f(n,n)2n−1(n∈Z+),Sna_{n}=\frac{\sqrt{f(n, n)}}{2^{n-1}}\left(n \in \mathbf{Z}_{+}\right), S_{n} be the sum of the first nn terms of the sequence {an}\left\{a_{n}\right\}, prove: Sn<6S_{n}<6.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

10. (1) From conditions (i) and (ii), we have
f(m,1)−f(1,1)=∑k=1m−1(f(k+1,1)−f(k,1))=2∑k=1m−1(k+1)⇒f(m,1)=m2+m−1⇒f(m,n)−f(m,1)=∑k=1n−1(f(m,k+1)−f(m,k))=2∑k=1n−1(m+k−1)⇒f(m,n)=(m+n)2−m−3n+1. \begin{array}{l} f(m, 1)-f(1,1) \\ =\sum_{k=1}^{m-1}(f(k+1,1)-f(k, 1)) \\ =2 \sum_{k=1}^{m-1}(k+1) \\ \Rightarrow f(m, 1)=m^{2}+m-1 \\ \Rightarrow f(m, n)-f(m, 1) \\ \quad=\sum_{k=1}^{n-1}(f(m, k+1)-f(m, k)) \\ \quad=2 \sum_{k=1}^{n-1}(m+k-1) \\ \Rightarrow f(m, n)=(m+n)^{2}-m-3 n+1 . \end{array}
(2) From (1), we get f(n,n)=(2n−1)2f(n, n)=(2 n-1)^{2}.

Then an=f(n,n)2n−1=(2n−1)(12)n−1a_{n}=\frac{\sqrt{f(n, n)}}{2^{n-1}}=(2 n-1)\left(\frac{1}{2}\right)^{n-1}.
Thus, Sn=∑k=1n(2k−1)(12)k−1S_{n}=\sum_{k=1}^{n}(2 k-1)\left(\frac{1}{2}\right)^{k-1},
12Sn=∑k=1n(2k−1)(12)k.  \frac{1}{2} S_{n}=\sum_{k=1}^{n}(2 k-1)\left(\frac{1}{2}\right)^{k} \text {. }

Subtracting the two equations, we get
12Sn=1+2∑k=1n−1(12)k−(2n−1)(12)n=3−12n−2−2n−12n. \begin{array}{l} \frac{1}{2} S_{n}=1+2 \sum_{k=1}^{n-1}\left(\frac{1}{2}\right)^{k}-(2 n-1)\left(\frac{1}{2}\right)^{n} \\ =3-\frac{1}{2^{n-2}}-\frac{2 n-1}{2^{n}} . \end{array}

Therefore, Sn=6−2n+32n−1<6S_{n}=6-\frac{2 n+3}{2^{n-1}}<6.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.