Let for x ∈ R, S₀(x) = x, Sₖ(x) = Cₖx + k ∫₀ˣ Sₖ₋₁(t)dt where C₀ = 1, Cₖ = 1 - ∫₀¹ Sₖ₋₁(x)dx, k = 1,

Q: Why can't I just find a general formula for $S_k(x)$ first?

While a general pattern might exist, JEE Advanced problems often test your ability to execute a multi-step process accurately. Trying to derive a complex general formula under pressure is a trap; the iterative approach is safer and more reliable. Focus on the immediate step.

Q: Do I really need to evaluate the integral every time?

Yes, because the constant $C_k$ is not a fixed value; it depends on the integral of the previous function $S_{k-1}(x)$. Each step is unique, and skipping the integration will lead to incorrect constants. Precision is your best friend here.

Q: Is there a shortcut for $S_2(3) + 6C_3$?

There is no magic shortcut here. The expression is designed to reward those who have carefully tracked their constants and functions. If you have calculated $S_2(x)$ and $C_3$ correctly, the final sum will simplify beautifully, which is a sign you are on the right track.

Question

Let for $x \in \mathbb{R}, S_0(x) = x, S_k(x) = C_kx + k \int_0^x S_{k-1}(t)dt$ where $C_0 = 1, C_k = 1 - \int_0^1 S_{k-1}(x)dx, k = 1, 2, 3, \dots$ Then $S_2(3) + 6C_3$ is equal to ————————.

Sigma Percentile · Accepted Answer

### The Recursive Dance: A Journey Through Iterative Calculus

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are embarking on a journey of recursive discovery.

When you first look at the definition $S_k(x) = C_kx + k \int_0^x S_{k-1}(t)dt$, it might feel like a labyrinth. But remember, every complex structure in mathematics is built from simple, foundational blocks.

We are going to build our tower of functions, one floor at a time, with the precision of an architect and the curiosity of an explorer.

### Phase 1: The Foundation

We start at the ground floor: $S_0(x) = x$. This is our base. It is simple, elegant, and the key to everything that follows.

Our first task is to find $C_1$. The formula tells us $C_k = 1 - \int_0^1 S_{k-1}(x)dx$. For $k=1$, this becomes:

$$C_1 = 1 - \int_0^1 S_0(x)dx$$

Substituting our base function, we get $C_1 = 1 - \int_0^1 x dx$. The integral of $x$ is $\frac{x^2}{2}$.

Evaluating this from $0$ to $1$ gives us $\frac{1}{2}$. Thus, $C_1 = 1 - \frac{1}{2} = \frac{1}{2}$. We have our first building block.

### Phase 2: The First Iteration

Now that we have $C_1$, we can construct $S_1(x)$. The recursive formula for $S_k(x)$ with $k=1$ is:

$$S_1(x) = C_1x + 1 \int_0^x S_0(t)dt$$

Substituting our known values, we have $S_1(x) = \frac{1}{2}x + \int_0^x t dt$. The integral of $t$ is $\frac{t^2}{2}$.

Evaluating this from $0$ to $x$ gives us $\frac{x^2}{2}$. So, $S_1(x) = \frac{1}{2}x + \frac{x^2}{2}$.

### Phase 3: The Second Iteration

We continue our ascent. To find $C_2$, we use $C_2 = 1 - \int_0^1 S_1(x)dx$. Substituting $S_1(x)$, we get:

$$C_2 = 1 - \int_0^1 \left( \frac{x}{2} + \frac{x^2}{2} \right) dx$$

The integral of $\frac{x}{2}$ is $\frac{x^2}{4}$, and the integral of $\frac{x^2}{2}$ is $\frac{x^3}{6}$. Evaluating from $0$ to $1$, we get $\frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12}$.

Therefore, $C_2 = 1 - \frac{5}{12} = \frac{7}{12}$. Now, we define $S_2(x)$ using $k=2$:

$$S_2(x) = C_2x + 2 \int_0^x S_1(t)dt$$

Substituting $C_2$ and $S_1(t)$, we get $S_2(x) = \frac{7}{12}x + 2 \int_0^x (\frac{t}{2} + \frac{t^2}{2}) dt$. The integral part becomes $2(\frac{t^2}{4} + \frac{t^3}{6}) = \frac{t^2}{2} + \frac{t^3}{3}$.

Evaluating from $0$ to $x$, we get $S_2(x) = \frac{7}{12}x + \frac{x^2}{2} + \frac{x^3}{3}$.

### Phase 4: The Final Stretch

We are almost at the summit. We need $C_3$ to complete our objective. $C_3 = 1 - \int_0^1 S_2(x)dx$. Substituting $S_2(x)$, we get:

$$C_3 = 1 - \int_0^1 \left( \frac{7}{12}x + \frac{x^2}{2} + \frac{x^3}{3} \right) dx$$

The integral is:

$$\left[ \frac{7x^2}{24} + \frac{x^3}{6} + \frac{x^4}{12} \right]_0^1 = \frac{7}{24} + \frac{1}{6} + \frac{1}{12} = \frac{7+4+2}{24} = \frac{13}{24}$$

Thus, $C_3 = 1 - \frac{13}{24} = \frac{11}{24}$. Now, the final calculation: $S_2(3) + 6C_3$.

First, $S_2(3) = \frac{7}{12}(3) + \frac{3^2}{2} + \frac{3^3}{3} = \frac{7}{4} + \frac{9}{2} + 9 = \frac{7+18+36}{4} = \frac{61}{4}$.

Then, $6C_3 = 6(\frac{11}{24}) = \frac{11}{4}$. Adding these together, $\frac{61}{4} + \frac{11}{4} = \frac{72}{4} = 18$.

The summit is reached! The final answer is **18**.

Animated Solution for Mathematics - Definite Integration: Let for $x \in R, S_{0} (x) = x, S_{k} (x) = C_{k} x + k \int_{0}^{x} S_{k - 1} (t) d t$ where $C_{0} = 1, C_{k} = 1 - \int_{0}^{1} S_{k - 1} (x) d x, k = 1, 2, 3, \dots$ Then $S_{2} (3) + 6 C_{3}$ is equal to ————————.

Enter Numerical Value:

Visualized Solution

Understanding the Recursive Definition

Setting up for $C_{1}$

Calculating the Constant $C_{1}$

Defining the Function $S_{1} (x)$

Computing $S_{1} (x)$

Setting up for $C_{2}$

Calculating the Constant $C_{2}$

Defining the Function $S_{2} (x)$

Computing $S_{2} (x)$

Setting up for $C_{3}$

Calculating the Constant $C_{3}$

Calculating $S_{2} (3)$

The Final Sum: $S_{2} (3) + 6 C_{3}$

Conclusion and Key Takeaways

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Recursive Dance

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

Understanding the Recursive Definition

Setting up for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Calculating the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​(x)

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​(x)

Setting up for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Calculating the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​(x)

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​(x)

Setting up for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C3​

Calculating the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C3​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​(3)

The Final Sum: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​(3)+6C3​