Animated Solution for Mathematics - Sequence and Series: If total number of runs scored in $n$ matches is $\frac{n+1}{4}(2^{n+1}-n-2)$ where $n>1$, and the runs scored in the $k^{th}$ match are given by $k\cdot 2^{n+1-k}$, where $1\le k\le n$. Find $n$.

Unraveling the AGP

Welcome, future engineer. Today, we are not just solving a problem; we are dissecting a mathematical structure. When you first look at the expression for the total runs, $\frac{n+1}{4}(2^{n+1}-n-2)$, it might feel intimidating.

It looks like a random collection of variables and powers. But in the world of JEE Advanced, intimidation is just a sign that you are about to learn something beautiful. Let us peel back the layers of this problem together.

The Fundamental Relation

We are given the total runs $S_n$ and the runs in the $k^{th}$ match $a_k=k\cdot 2^{n+1-k}$. The bridge between these two is the concept of summation. If you have a series of matches, the total runs are simply the sum of runs in each match.

Mathematically, this is our bedrock:

Before we panic, let us simplify the exponent. We know that $2^{n+1-k}=\frac{2^{n+1}}{2^k}$. Since $2^{n+1}$ does not depend on $k$, it is a constant relative to our summation.

We can pull it out:

Suddenly, the problem transforms. We are no longer looking at a complex formula; we are looking at a clean, elegant summation: $S={\sum }_{k=1}^n\frac{k}{2^k}$.

The AGP Revelation

Let us write out the first few terms of $S$:

Look closely at the numerators: $1,2,3,\dots ,n$. That is an Arithmetic Progression (AP) with a common difference of $1$.

Now look at the denominators: $2^1,2^2,2^3,\dots ,2^n$. That is a Geometric Progression (GP) with a common ratio of $\frac{1}{2}$. When an AP and a GP are multiplied term-by-term, we get an Arithmetico-Geometric Progression (AGP).

This is a classic pattern in competitive mathematics. Recognizing it is half the battle won.

The Art of the Shift

How do we solve an AGP? We use the 'Shift and Subtract' method. We multiply the entire series $S$ by the common ratio of the GP, which is $r=\frac{1}{2}$.

Now, we align this new series under the original $S$, shifting it by one position to the right so that terms with the same power of $2$ in the denominator line up:

Subtracting the second from the first, we get:

The Elegant Cancellation

The bracketed term is now a simple Geometric Progression! The sum of a GP is $\frac{a(1-r^n)}{1-r}$. Here, $a=\frac{1}{2}$ and $r=\frac{1}{2}$.

Since $1-\frac{1}{2}=\frac{1}{2}$, the fraction simplifies beautifully to $1-\frac{1}{2^n}$. Thus:

Multiplying by $2$, we find $S=2-\frac{1}{2^{n-1}}-\frac{n}{2^n}$. With a common denominator of $2^n$, this becomes $S=\frac{2^{n+1}-2-n}{2^n}$.

Now, we bring this back to our original equation: $S_n=2^{n+1}\cdot S$. Substituting our result for $S$:

Since $\frac{2^{n+1}}{2^n}=2$, we get $S_n=2(2^{n+1}-n-2)$. Equating this to the given total runs $\frac{n+1}{4}(2^{n+1}-n-2)$, we see the common factor $(2^{n+1}-n-2)$ on both sides.

Since $n>1$, this term is non-zero and cancels out perfectly:

And there it is. The complexity collapses into a simple integer. Remember, in physics and mathematics, complexity is often just a mask for a simpler, underlying truth.