Let S_n = 1 + q + q² + .... + qⁿ and T_n = 1 + ((q+1)/2) + ((q+1)/2)² + .... + ((q+1)/2)ⁿ where q is

Q: Wait, why does $S_n$ have $n+1$ terms? Shouldn't it just be $n$ terms?

That is a classic trap! Look closely at the sequence: $1 + q + q^2 + \dots + q^n$. The powers of $q$ range from $0$ to $n$. Counting from $0$ to $n$ gives us $n+1$ terms, not $n$. Always count the number of terms by looking at the exponents, not just the last index.

Q: Why do we subtract the $r=0$ term when evaluating the binomial sum?

Great observation. The standard binomial expansion $\sum_{r=0}^{n} {}^{n}C_r x^r$ includes the $r=0$ term, which is ${}^{n}C_0 x^0 = 1$. Our summation starts from $r=1$. Therefore, to use the standard formula, we must take the full sum and subtract the $r=0$ term that we don't need.

Q: Is there a faster way to solve this without all the algebraic manipulation?

While you could potentially use induction for specific values of $n$, the algebraic approach is the most robust and elegant method for JEE Advanced. It builds your intuition for manipulating series, which is a vital skill for higher-level calculus and combinatorics.

Question

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 $q 
eq 1$. If $^{101}C_1 + ^{101}C_2 \cdot S_1 + \dots + ^{101}C_{101} \cdot S_{100} = \alpha T_{100}$ then $\alpha$ is equal to:

Options:

(A) 200

(B) $2^{100}$

(C) $2^{99}$

(D) 202

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a mathematical onion to reveal a hidden symmetry. We are given two sequences, $S_n$ and $T_n$, and a complex-looking summation that seems designed to intimidate.

But fear not, for beneath the surface lies a beautiful, structured dance of algebra.

### Decoding the Sequences

First, let us look at $S_n = 1 + q + q^2 + \dots + q^n$. This is a classic Geometric Progression where the first term $a = 1$ and the common ratio is $q$. Since the powers of $q$ range from $0$ to $n$, there are exactly $n+1$ terms.

Using the sum formula for a G.P., we obtain:

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

Now, consider $T_n = 1 + (\frac{q+1}{2}) + (\frac{q+1}{2})^2 + \dots + (\frac{q+1}{2})^n$. This is also a G.P., but with a common ratio $r = \frac{q+1}{2}$. Applying the same logic, we find:

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

### The Sigma Transformation

The given series is $^{101}C_1 + ^{101}C_2 S_1 + \dots + ^{101}C_{101} S_{100}$. We can express this using Sigma notation as $\sum_{r=1}^{101} {}^{101}C_r S_{r-1}$.

Notice how the subscript of $S$ is always one less than the index of $C$. Substituting our formula for $S_{r-1}$, we get:

$$\sum_{r=1}^{101} {}^{101}C_r \frac{q^r-1}{q-1}$$

Since $q-1$ is independent of $r$, we pull it out of the summation:

$$\frac{1}{q-1} \sum_{r=1}^{101} {}^{101}C_r (q^r - 1)$$

### The Binomial Dance

Now, we distribute the binomial coefficient to split the sum into two distinct parts:

$$\frac{1}{q-1} \left[ \sum_{r=1}^{101} {}^{101}C_r q^r - \sum_{r=1}^{101} {}^{101}C_r \right]$$

The first part is almost the binomial expansion of $(1+q)^{101}$. The only missing piece is the $r=0$ term, which is ${}^{101}C_0 q^0 = 1$. Thus, the first sum is $(1+q)^{101} - 1$.

The second part is the sum of binomial coefficients, which is $2^{101} - 1$ (again, subtracting the $r=0$ term). Combining these, our LHS becomes:

$$\frac{(1+q)^{101} - 1 - (2^{101} - 1)}{q-1} = \frac{(1+q)^{101} - 2^{101}}{q-1}$$

### The Grand Cancellation

Finally, we look at the RHS, $\alpha T_{100}$. Substituting $n=100$ into our $T_n$ formula, we get:

$$T_{100} = \frac{(\frac{q+1}{2})^{101}-1}{\frac{q+1}{2}-1} = \frac{(q+1)^{101} - 2^{101}}{2^{101}} \cdot \frac{2}{q-1} = \frac{(q+1)^{101} - 2^{101}}{2^{100}(q-1)}$$

Equating the LHS and RHS, we have:

$$\frac{(1+q)^{101} - 2^{101}}{q-1} = \alpha \frac{(q+1)^{101} - 2^{101}}{2^{100}(q-1)}$$

The massive term involving $q$ cancels out perfectly, leaving $1 = \frac{\alpha}{2^{100}}$. Therefore, the final result is:

$$\mathbf{\alpha = 2^{100}}$$

Select Answer:

Visualized Solution

Analyze the sequence $S_{n}$

Analyze the sequence $T_{n}$

Express the given series in Sigma notation

Substitute $S_{r - 1}$ into the summation

Split the summation

Evaluate the first binomial sum

Evaluate the second binomial sum

Simplify the Left Hand Side (LHS)

Calculate $T_{100}$

Simplify the expression for $T_{100}$

Equate LHS and RHS to find $α$

Final Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

Decoding the Sequences

The Sigma Transformation

The Binomial Dance

The Grand Cancellation

Similar Questions

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

Let $n \in N$ and $[x]$ denote the greatest integer less than or equal to $x$ . If the sum of $(n + 1)$ terms $^{n} C_{0}, 3 \cdot^{n} C_{1}, 5 \cdot^{n} C_{2}, 7 \cdot^{n} C_{3}, \dots$ is equal to $2^{100} \cdot 101$ , then $2 [\frac{n - 1}{2}]$ is equal to ___

If $(\frac{1}{^{15} C _{0}} + \frac{1}{^{15} C _{1}}) (\frac{1}{^{15} C _{1}} + \frac{1}{^{15} C _{2}}) \dots (\frac{1}{^{15} C _{12}} + \frac{1}{^{15} C _{13}}) = \frac{α ^{13}}{^{14} C _{0} \cdot ^{14} C _{1} \dots ^{14} C _{12}}$ , then $30 α$ is equal to .........

If $1 + (^{49} C_{0} +^{49} C_{1} + \dots +^{49} C_{49} + 1) (^{50} C_{2} +^{50} C_{4} + \dots +^{50} C_{50})$ is equal to $2^{n} \cdot m$ , where $m$ is odd, then $n + m$ is equal to ______.

If $\sum_{r = 0}^{10} (\frac{1 0 ^{r + 1} - 1}{1 0 ^{r}}) \cdot^{11} C_{r + 1} = \frac{α ^{11} - 1 1 ^{11}}{1 0 ^{10}}$ , then $α$ is equal to :

The value of $\frac{^{100} C _{50}}{51} + \frac{^{100} C _{51}}{52} + \dots + \frac{^{100} C _{100}}{101}$ is :

If $S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}}$ and $t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C _{r}}$ , then $\frac{t _{n}}{S _{n}}$ is equal to

If $C_{r} =^{25} C_{r}$ and $C_{0} + 5 \cdot C_{1} + 9 \cdot C_{2} + \dots + 101 \cdot C_{25} = 2^{25} \cdot k$ then $k$ is equal to__________.

$\sum_{i, j = 0 i \neq = j}^{n}^{n} C_{i}^{n} C_{j}$ is equal to

$\sum_{i = 1}^{20} \frac{^{20} C _{i - 1}}{^{20} C _{i} + ^{20} C _{i - 1}} = \frac{k}{21}$ , then k equals :

Select Answer:

Visualized Solution

Analyze the sequence .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​

Analyze the sequence .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tn​

Express the given series in Sigma notation

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sr−1​ into the summation

Split the summation

Evaluate the first binomial sum

Evaluate the second binomial sum

Simplify the Left Hand Side (LHS)

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T100​

Simplify the expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T100​

Equate LHS and RHS to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α