Animated Solution for Mathematics - Differentiation: Let $f(x)={\sum }_{k=1}^{10}k\cdot x^k,x\in \mathbb{R}$, if $2f(2)+f^{\prime }(2)=119(2{)}^n+1$ then $n$ is equal to ______.

Welcome, future engineer. Today, we are going to dismantle a problem that looks like a monster but is actually a masterpiece of algebraic symmetry. We are looking at the function defined by:

At first glance, this is a summation, a terrifying wall of terms. However, notice that the coefficients $1,2,3,\dots ,10$ are in an arithmetic progression, and the powers $x,x^2,x^3,\dots ,x^{10}$ are in a geometric progression.

This is an Arithmetico-Geometric Progression (AGP). In the JEE Advanced, whenever you see this structure, do not panic. We have a standard, elegant weapon for this: the shift-and-subtract method.

Imagine you are standing on a bridge. To get to the other side, you need to simplify the expression. We define the sum as:

Now, we multiply this entire series by the common ratio, which is $x$. This gives us:

Here is the crucial moment. We align these two equations and subtract $xS$ from $S$:

Look at the magic! The middle terms collapse into a simple geometric progression:

This is the moment where the complexity vanishes. We can now use the standard sum formula for a GP, $\frac{a(1-r^n)}{1-r}$, to simplify the expression:

With the GP sum in hand, our function becomes:

This is our master equation. Now, we need $f^{\prime }(x)$. I know, differentiating a quotient looks intimidating, but take a deep breath.

We apply the quotient rule: $\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$. It is a test of your algebraic patience, not your intelligence. As we differentiate and substitute $x=2$, the denominator $(1-x{)}^2$ becomes $(-1{)}^2=1$, which simplifies our lives immensely.

After careful calculation, we find:

We are in the endgame now. We have $f(2)=2+18\cdot 2^{10}$ and $f^{\prime }(2)=83\cdot 2^{10}-3$. The problem asks us to evaluate $2f(2)+f^{\prime }(2)$.

Substituting our values, we get:

Expanding this, we get:

Combining the terms, we arrive at:

Comparing this to the given form $119\cdot 2^n+1$, it is crystal clear that $n=10$. You have successfully navigated the complexity and arrived at the truth. This is the essence of JEE—not just solving, but seeing the structure beneath the surface.