Given sn = 1 + q + q² + .... + qⁿ; Sn = 1 + (q+1)/2 + ((q+1)/2)² + .... + ((q+1)/2)ⁿ, q ≠ 1 Prove th

Q: Why can't we just use the sum of G.P. formula directly on the whole LHS?

The LHS is not a simple G.P. It is a weighted sum where the weights are binomial coefficients. You cannot treat the entire expression as a single G.P. because the coefficients ${^{n+1}C_k}$ change with every term. You must treat the binomial coefficients and the G.P. terms as separate entities interacting through the summation operator.

Q: Why does the index shift from $k$ to $r$ matter?

It is a psychological and algebraic necessity. By shifting to $r=k+1$, you align the index with the power of $q$, making the expression identical to the standard binomial expansion $(1+q)^{n+1}$. It transforms a confusing summation into a recognizable, standard form.

Q: What if $q=1$?

The problem explicitly states $q \neq 1$. If $q=1$, the formula for the sum of a G.P. involves division by zero. This is a singularity where the standard formula breaks down, requiring a different approach (simply counting the terms), but for this proof, we are safely within the domain where the formula holds.

Question

Given $s_n = 1 + q + q^2 + .... + q^n$; $S_n = 1 + \frac{q+1}{2} + (\frac{q+1}{2})^2 + .... + (\frac{q+1}{2})^n, q 
e 1$ Prove that $^{n+1}C_1 + ^{n+1}C_2s_1 + ^{n+1}C_3s_2 + .... + ^{n+1}C_{n+1}s_n = 2^n S_n$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a problem that might look like a chaotic jumble of symbols, but it is a beautifully choreographed dance of series and binomial coefficients.

When you first look at the expression $^{n+1}C_1 + ^{n+1}C_2s_1 + ^{n+1}C_3s_2 + .... + ^{n+1}C_{n+1}s_n = 2^n S_n$, it is natural to feel overwhelmed. In JEE Advanced, the secret is never to attack the problem head-on with brute force. Instead, we look for the hidden structure and the patterns woven into the fabric of the equation.

### Phase 1: Deconstructing the Building Blocks

Before we touch the main equation, let us understand our components. We are given the series $s_n = 1 + q + q^2 + .... + q^n$.

This is a classic Geometric Progression (G.P.) with $n+1$ terms and a common ratio of $q$. We know the sum of a G.P. is given by the formula $S = \frac{a(1-r^N)}{1-r}$.

Applying this to our general term $s_k$, we get:

$$s_k = \frac{1-q^{k+1}}{1-q}$$

This is our first weapon. We have condensed a sprawling series into a single, elegant fraction. Keep this in your toolkit; we will need it soon.

### Phase 2: The Sigma Transformation

Now, let us gaze upon the Left Hand Side (LHS). It is a long, intimidating chain of terms. Notice the pattern: the subscript of the combination term $C$ is always exactly one more than the subscript of the series term $s$.

Whenever you see a pattern like this, Sigma notation ($\sum$) is your best friend. We can rewrite the entire LHS as:

$$\sum_{k=0}^n {^{n+1}C_{k+1}} s_k$$

Suddenly, the chaos is organized. Substitute our closed form of $s_k$ into this summation. Since $\frac{1}{1-q}$ is independent of $k$, we can pull it outside the summation:

$$\frac{1}{1-q} \sum_{k=0}^n {^{n+1}C_{k+1}} (1-q^{k+1})$$

### Phase 3: The Binomial Bridge

We are now looking at the expression:

$$\frac{1}{1-q} \left[ \sum_{k=0}^n {^{n+1}C_{k+1}} (1) - \sum_{k=0}^n {^{n+1}C_{k+1}} q^{k+1} \right]$$

To make these look like the standard Binomial expansion $(1+x)^N = \sum_{r=0}^N {^NC_r} x^r$, we perform a variable shift. Let $r = k+1$. As $k$ ranges from $0$ to $n$, $r$ ranges from $1$ to $n+1$.

Our expression transforms into:

$$\frac{1}{1-q} \left[ \sum_{r=1}^{n+1} {^{n+1}C_r} - \sum_{r=1}^{n+1} {^{n+1}C_r} q^r \right]$$

Recall that the sum of binomial coefficients $\sum_{r=1}^N {^NC_r} = 2^N - 1$, and the sum $\sum_{r=1}^N {^NC_r} x^r = (1+x)^N - 1$. Applying these identities, the LHS simplifies to:

$$\frac{(2^{n+1} - 1) - ((1+q)^{n+1} - 1)}{1-q} = \frac{2^{n+1} - (1+q)^{n+1}}{1-q}$$

### Phase 4: The Mirror Image

Now, we turn our attention to the Right Hand Side (RHS): $2^n S_n$. We know $S_n$ is another G.P. with common ratio $\frac{q+1}{2}$. Using the G.P. sum formula:

$$S_n = \frac{1 - (\frac{q+1}{2})^{n+1}}{1 - \frac{q+1}{2}}$$

Simplifying the denominator $1 - \frac{q+1}{2}$ gives us $\frac{1-q}{2}$. When we multiply this by $2^n$, the algebra aligns:

$$2^n \cdot \frac{1 - \frac{(1+q)^{n+1}}{2^{n+1}}}{\frac{1-q}{2}} = 2^n \cdot \frac{2^{n+1} - (1+q)^{n+1}}{2^{n+1}} \cdot \frac{2}{1-q}$$

The $2^n$ and the $2$ in the numerator cancel the $2^{n+1}$ in the denominator, leaving us with the exact same expression:

$$\frac{2^{n+1} - (1+q)^{n+1}}{1-q}$$

### Conclusion: The Beauty of Symmetry

We have arrived. The LHS equals the RHS. This problem was never about brute force calculation; it was about recognizing the underlying symmetry of the Binomial Theorem and the Geometric Progression.

When you see these structures in the future, do not panic. Break them down, use Sigma notation to organize your thoughts, and trust the identities. You have mastered this proof, and in doing so, you have mastered a piece of the beautiful, logical architecture of mathematics.

Visualized Solution

Analyzing the Given Series $s_{n}$ and $S_{n}$

Formula for the Sum of G.P.

Condensing the Left Hand Side

Substituting the G.P. Sum into LHS

Expanding the Summation

Standard Binomial Sums

Simplifying the Left Hand Side

Expanding the Right Hand Side

Simplifying the RHS Denominator

Simplifying the RHS Numerator

Final Proof: LHS = RHS

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

Given that $C_{1} + 2 C_{2} x + 3 C_{3} x^{2} + .... + 2 n C_{2 n} x^{2 n - 1} = 2 n (1 + x)^{2 n - 1}$ where $C_{r} = \frac{( 2 n )!}{r ! ( 2 n - r )!}, r = 0, 1, 2, ...., 2 n$ . Prove that $C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - .... - 2 n C_{2 n}^{2} = (- 1)^{n} n^{n} C_{n}$ .

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + .... + C_{n} x^{n}$ then show that the sum of the products of the $C_{i}$ 's taken two at a time, represented by $\sum_{0 \leq i < j \leq n} C_{i} C_{j}$ is equal to $2^{2 n - 1} - \frac{( 2 n )!}{2 ( n ! ) ^{2}}$

Prove that $C_{0} - 2^{2} C_{1} + 3^{2} C_{2} - .... + (- 1)^{n} (n + 1)^{2} C_{n} = 0, n > 2$ , where $C_{r} =^{n} C_{r}$ .

Let $n$ be a positive integer and $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{2 n} x^{2 n}$ . Show that $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

If $n \geq 2$ is a positive integer, then the sum of the series $^{n + 1} C_{2} + 2 (^{2} C_{2} +^{3} C_{2} +^{4} C_{2} + ... +^{n} C_{2})$ is :

Prove that $2^{k} (0 n) (k n) - 2^{k - 1} (1 n) (k - 1 n - 1) + 2^{k - 2} (2 n) (k - 2 n - 2) - .... + (- 1)^{k} (k n) (0 n - k) = (k n)$

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

Let $S_{n} = 1 + q + q^{2} + \dots + q^{n}$ and $T_{n} = 1 + (\frac{q + 1}{2}) + (\frac{q + 1}{2})^{2} + \dots + (\frac{q + 1}{2})^{n}$ where $q$ is a real number and $q \neq = 1$ . If $^{101} C_{1} +^{101} C_{2} \cdot S_{1} + \dots +^{101} C_{101} \cdot S_{100} = α T_{100}$ then $α$ is equal to:

For any positive integer $m, n$ (with $n \geq m$ ), let $(m n) =^{n} C_{m}$ . Prove that $(m n) + (m n - 1) + (m n - 2) + .... + (m m) = (m + 1 n + 1)$ . Hence or otherwise, prove that $(m n) + 2 (m n - 1) + 3 (m n - 2) + .... + (n - m + 1) (m m) = (m + 2 n + 2)$ .