JEE Mathematics Solution: Binomial Theorem and Geometric Progressions

Key Concepts Used

This problem elegantly combines concepts from two fundamental areas of mathematics:

1. Geometric Progression (GP) Sum: A geometric progression is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The sum of the first $N$ terms of a GP with first term $a$ and common ratio $r$ is given by $S_N = \frac{a(1-r^N)}{1-r}$, provided $r \neq 1$.
2. Binomial Theorem: This theorem describes the algebraic expansion of powers of a binomial $(x+y)^n$. The general formula is $(x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$, where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ are the binomial coefficients. Key identities derived from this include:
   • $\sum_{k=0}^{n} \binom{n}{k} = 2^n$ (by setting $x=1, y=1$)
   • $\sum_{k=0}^{n} \binom{n}{k} y^k = (1+y)^n$ (by setting $x=1$)

Understanding the Series $S_n$ and $T_n$

Let's first express the given series $S_n$ and $T_n$ in their closed forms using the GP sum formula.

• Series $S_n$: $S_n = 1 + q + q^2 + \dots + q^n$ This is a Geometric Progression with the first term $a = 1$, common ratio $r = q$, and a total of $(n+1)$ terms. Using the GP sum formula for $(n+1)$ terms: $S_n = \frac{1 \cdot (1 - q^{n+1})}{1 - q} = \frac{1 - q^{n+1}}{1 - q} \quad \text{ (since } q \ne 1 \text{)}$

• Series $T_n$: $T_n = 1 + \left( \frac{q+1}{2} \right) + \left( \frac{q+1}{2} \right)^2 + \dots + \left( \frac{q+1}{2} \right)^n$ This is also a Geometric Progression with the first term $a = 1$, common ratio $r = \frac{q+1}{2}$, and a total of $(n+1)$ terms. Using the GP sum formula for $(n+1)$ terms: $T_n = \frac{1 \cdot \left(1 - \left(\frac{q+1}{2}\right)^{n+1}\right)}{1 - \frac{q+1}{2}}$ Simplifying the denominator: $1 - \frac{q+1}{2} = \frac{2 - (q+1)}{2} = \frac{2 - q - 1}{2} = \frac{1-q}{2}$. Substituting this back into the formula for $T_n$: $T_n = \frac{1 - \left(\frac{q+1}{2}\right)^{n+1}}{\frac{1-q}{2}} = \frac{2 \left(1 - \left(\frac{q+1}{2}\right)^{n+1}\right)}{1-q}$

Analyzing the Given Equation

The problem provides the following equation: $\binom{101}{1} + \binom{101}{2} S_1 + \dots + \binom{101}{101} S_{100} = \alpha T_{100}$ We can write the Left Hand Side (LHS) as a summation: $\text{LHS} = \sum_{k=1}^{101} \binom{101}{k} S_{k-1}$ Now, we substitute the closed form of $S_{k-1} = \frac{1 - q^{(k-1)+1}}{1-q} = \frac{1 - q^k}{1-q}$ into the LHS: $\text{LHS} = \sum_{k=1}^{101} \binom{101}{k} \left( \frac{1 - q^k}{1-q} \right)$ And the Right Hand Side (RHS) is $\alpha T_{100}$. Substituting the closed form of $T_{100}$: $\text{RHS} = \alpha \cdot \frac{2 \left(1 - \left(\frac{q+1}{2}\right)^{101}\right)}{1-q}$ So the main equation becomes: $\sum_{k=1}^{101} \binom{101}{k} \left( \frac{1 - q^k}{1-q} \right) = \alpha \cdot \frac{2 \left(1 - \left(\frac{q+1}{2}\right)^{101}\right)}{1-q}$

Simplifying the Equation

Since $q \ne 1$, we know that $(1-q) \ne 0$. Therefore, we can multiply both sides of the equation by $(1-q)$ to simplify: $\sum_{k=1}^{101} \binom{101}{k} (1 - q^k) = 2\alpha \left(1 - \left(\frac{q+1}{2}\right)^{101}\right)$ Let's simplify the LHS further: $\text{LHS} = \sum_{k=1}^{101} \left( \binom{101}{k} - \binom{101}{k} q^k \right)$ $\text{LHS} = \sum_{k=1}^{101} \binom{101}{k} - \sum_{k=1}^{101} \binom{101}{k} q^k$ Now, we use the binomial identities:

• $\sum_{k=0}^{n} \binom{n}{k} = 2^n$. So, $\sum_{k=1}^{101} \binom{101}{k} = \left( \sum_{k=0}^{101} \binom{101}{k} \right) - \binom{101}{0} = 2^{101} - 1$.
• $\sum_{k=0}^{n} \binom{n}{k} x^k = (1+x)^n$. So, $\sum_{k=1}^{101} \binom{101}{k} q^k = \left( \sum_{k=0}^{101} \binom{101}{k} q^k \right) - \binom{101}{0} q^0 = (1+q)^{101} - 1$.

Substitute these back into the LHS: $\text{LHS} = (2^{101} - 1) - ((1+q)^{101} - 1)$ $\text{LHS} = 2^{101} - 1 - (1+q)^{101} + 1$ $\text{LHS} = 2^{101} - (1+q)^{101}$ Now, let's look at the RHS: $\text{RHS} = 2\alpha \left(1 - \left(\frac{1+q}{2}\right)^{101}\right)$ We can rewrite the term inside the parenthesis: $1 - \left(\frac{1+q}{2}\right)^{101} = 1 - \frac{(1+q)^{101}}{2^{101}} = \frac{2^{101} - (1+q)^{101}}{2^{101}}$ So, the RHS becomes: $\text{RHS} = 2\alpha \left( \frac{2^{101} - (1+q)^{101}}{2^{101}} \right)$

Equating LHS and RHS to find $\alpha$

Now we equate the simplified LHS and RHS: $2^{101} - (1+q)^{101} = 2\alpha \left( \frac{2^{101} - (1+q)^{101}}{2^{101}} \right)$ We need to consider the term $2^{101} - (1+q)^{101}$. If $q=1$, then $2^{101} - (1+1)^{101} = 2^{101} - 2^{101} = 0$. But the problem states $q \ne 1$. Thus, $2^{101} - (1+q)^{101}$ is generally non-zero. If this term were zero, it would imply $1+q=2$, meaning $q=1$, which is excluded. Therefore, we can safely divide both sides by $2^{101} - (1+q)^{101}$: $1 = \frac{2\alpha}{2^{101}}$ Now, solve for $\alpha$: $2^{101} = 2\alpha$ $\alpha = \frac{2^{101}}{2}$ $\alpha = 2^{100}$

Tips and Common Mistakes to Avoid

• Index Management in GP Sums: Be careful with the number of terms. If a series goes from $q^0$ to $q^n$, there are $n+1$ terms, not $n$. Incorrect term count is a frequent source of error.
• Binomial Sum Limits: Remember that standard binomial identities like $\sum_{k=0}^n \binom{n}{k}$ start from $k=0$. If your sum starts from $k=1$, you must subtract the $k=0$ term (i.e., $\binom{n}{0}$).
• Algebraic Simplification: Pay close attention to signs and denominators, especially when combining fractions or terms with powers.
• Assumptions: Note any conditions given in the problem, like $q \ne 1$. These conditions are often crucial for safely performing operations like division.

Summary

By first expressing the given series $S_n$ and $T_n$ in their closed forms using the Geometric Progression sum formula, the complex summation on the LHS was simplified using key binomial identities. Comparing the simplified LHS with the $\alpha T_{100}$ term on the RHS allowed for direct calculation of $\alpha$. The final result is $\alpha = 2^{100}$. This problem demonstrates the power of recognizing underlying series structures and applying standard summation formulas to simplify complex expressions.