For any arrangement of $99$ weights, we can obtain a corresponding arrangement of $100$ weights by putting the $100$ on the pan that $50$ is currently on and moving the $50$ to the other pan, so we have already obtained $T(99)$ arrangements. So we simply need to determine an arrangement where $100$ and $50$ resides on the same pan to prove that $T(100) > T(99)$. This is easy: one example is $100,98,96,\ldots,2$ without $26$ and with $1$, versus $99,97,95,\ldots,3$ with $26$.

shivangjindal wrote: For a natural number $n$, let $T(n)$ denote the number of ways we can place $n$ objects of weights $1,2,\cdots, n$ on a balance such that the sum of the weights in each pan is the same. Prove that $T(100) > T(99)$. Is it $T$ an increasing function?

No; $T(n) = 0$ for all $n \equiv 1,2 \pmod 4$, and $T(n) > 0$ otherwise. I'm pretty sure $T(n)$ is increasing among the nonzero values though. EDIT: $T(3) = T(4)$. But otherwise $T(4n-1) < T(4n)$ for $n \ge 2$; the proof is similar as above, with simply figuring out the remaining weighing. Now it remains to prove that $T(4n) < T(4n+3)$...

Okay. $T(4n+1) = T(4n+2) = 0$, and $T(3) = T(4) = 2$ (1,2 vs 3 and 3 vs 1,2; 1,4 vs 2,3 and 2,3 vs 1,4) Now, $T(4n-1) < T(4n)$. Take an arrangement of $4n$ weights, move the weight $2n$ to the other pan and place $4n$ to the old pan containing $2n$; this shows that $T(4n) \ge T(4n-1)$. Then, observe the following: place $4n, 2n$ on the left pan, $3n, n$ on the right pan. Place $1,2,\ldots,n-1, 3n+1,3n+2,\ldots,4n-1$ on the right pan and $n+1,n+2,\ldots,3n$ on the left pan. It can be easily verified that the left pan is heavier by $2n$, so exchange $1$ and $n+1$. This obtains an arrangement where $4n$ and $2n$ are on the same pan and thus can't possibly come from an arrangement of $4n-1$ weights, thus $T(4n)$ counts at least one more arrangement than $T(4n-1)$, enough to prove that $T(4n) > T(4n-1)$. Also, $T(4n) < T(4n+3)$. Similar as above. Take an arrangement of $4n$ weights, move weight $2n$ to the other pan and put $4n+3$ along it, and put $4n+1, 4n+2$ on the opposite pan. This proves that $T(4n+3) \ge T(4n)$. For the remaining arrangement, put $4n+1, 4n+3$ on the left pan, $4n+2$ on the right pan, $1,2,\ldots,n,3n+1,3n+2,\ldots,4n$ on the right pan, $n+1,n+2,\ldots,3n$ on the left pan. It can be verified that the left pan is heavier by $4n+2$, so move $2n+1$ to the right pan. Here we obtain an arrangement where $4n+1$ and $4n+2$ are on different pans, so it cannot possibly come from an arrangement of $4n$ weights and thus $T(4n+3) > T(4n)$.

There is another idea which numerically doesn't look much feasible though. Obviously $T(100)$ is the coefficient of $x^{2525}$ in the expansion of $(1+x)(1+x^2)(1+x^3) \cdots (1+x^{100})$ And $T(99)$ the coefficient of $x^{2475}$ in the expansion of $(1+x)(1+x^2)(1+x^3) \cdots (1+x^{99})$ Is there some way in which this approach can work ?

chaotic_iak wrote: For any arrangement of $99$ weights, we can obtain a corresponding arrangement of $100$ weights by putting the $100$ on the pan that $50$ is currently on and moving the $50$ to the other pan, so we have already obtained $T(99)$ arrangements. So we simply need to determine an arrangement where $100$ and $50$ resides on the same pan to prove that $T(100) > T(99)$. This is easy: one example is $100,98,96,\ldots,2$ without $26$ and with $1$, versus $99,97,95,\ldots,3$ with $26$. Here is there some method for determining the example? Because in exam I guess we won't get full marks for writing directly the example.

I know it, you can derive a formula such that a+b=c+d and a,b,c,d are distinct and are from 1,2,3,..n

utkarshgupta wrote: There is another idea which numerically doesn't look much feasible though. Obviously $T(100)$ is the coefficient of $x^{2525}$ in the expansion of $(1+x)(1+x^2)(1+x^3) \cdots (1+x^{100})$ And $T(99)$ the coefficient of $x^{2475}$ in the expansion of $(1+x)(1+x^2)(1+x^3) \cdots (1+x^{99})$ Is there some way in which this approach can work ? How did you get it?? Can anyone help me.

It's Generating functions if I am not wrong. You may refer to Pathfinder for it. It's quite a famous way of counting things, we are taught this every year at RMO workshop at Allen

Let $\chi(n)$ denote the number of ways in which we can pick up subsets from a set of $\{1,2,\cdots,n\}$ such that the sum of elements of each subset $=\frac{n(n+1)}{4}\in\Bbb{N}$. Now, we can reframe the question as just proving that $\chi(100)>\chi(99)$.. Define $f:\chi(99)\rightarrow \chi(100)$ as a function defined by : if $C$ is an element in the domain of $f$ such that $50\in C$ then $f(C)=\{100\}\cup (C-\{50\})$ or else if $50\notin C$ then $f(C)=C\cup \{50\}$. Clearly, if $f(X)=f(Y)$ then $X=Y$ and therefore, $f$ is an one-one function. Observe that $f$ is so defined that no element in the range of $f$ can contain both $100$ and $50$ together.. But, if we choose a set, $\{(1,100),(2,99),(3,98),\cdots,(24,77),(50,51)\}$ we observe that it has both $100$ and $50$ together.. Note Here, $\{(a,b)\}$ simply means $\{a,b\}$ and has no other special meaning.. The first brackets are used only for the sake of convenience of grouping the terms, in an attempt to ease the explanation.. We also observe, that the sum of elements of this set is $25.101=2525$ thus it must belong to $\chi(100)$. But it is not in the range of $f$. Therefore, $f$ is not an onto function.. This implies, $n(\chi(100))>n(\chi(99))$ Notation here, $n(A)$ denotes the cardinality of some set $A$ Thus $T(100)>T(99)$ Q.E.D. $\square$

nice solution!