Evaluate Alternating Series Using Telescoping Method

Learn how to evaluate a complex alternating series using pattern recognition and telescoping method. This approach helps simplify series problems in..

❓ Question

Evaluate the sum of the series:

S = 2 \times 1 \times (\binom{20}{4}) - 3 \times 2 \times (\binom{20}{5}) + 4 \times 3 \times (\binom{20}{6}) - 5 \times 4 \times (\binom{20}{7}) + \dots + 18 \times 17 \times (\binom{20}{20}) .

🖼️ Question Image

The sum of the series 2 × 1 × ²⁰C₄ − 3 × 2 × ²⁰C₅ + 4 × 3 × ²⁰C₆ − 5 × 4 × ²⁰C₇ + ⋯ + 18 × 17 × ²⁰C₂₀, is equal to:

✍️ Short Explanation

This problem is based on:

👉 Binomial theorem
👉 Series transformation
👉 Combination identities.

Main idea:

Convert terms into standard summation form and use:

r(r-1)\binom{20}{r} = 20\cdot19\binom{18}{r-2}

🔷 Step 1 — Write General Term 💯

Observe pattern:

2\cdot1\binom{20}{4} - 3\cdot2\binom{20}{5} + 4\cdot3\binom{20}{6} -\cdots

For general term:

T_r = (-1)^r(r-2)(r-3)\binom{20}{r}

where:

r=4 \text{ to } 20

Thus:

S= \sum_{r=4}^{20} (-1)^r(r-2)(r-3)\binom{20}{r}

Now:

(r-2)(r-3)=r^2-5r+6

But easier method is factorial identity.

🔷 Step 2 — Use Combination Identity

We know:

r(r-1)\binom{20}{r} = 20\cdot19\binom{18}{r-2}

Now:

(r-2)(r-3)\binom{20}{r} = 20\cdot19\binom{18}{r-2} -4(r-1)\binom{20}{r} +6\binom{20}{r}

Instead of lengthy expansion, shift index directly.

Let:

k=r-2

Then:

S= \sum_{k=2}^{18} (-1)^k\,k(k-1)\binom{20}{k+2}

Now use identity:

k(k-1)\binom{20}{k+2} = 20\cdot19\binom{18}{k}

Hence:

S= 20\cdot19 \sum_{k=2}^{18} (-1)^k\binom{18}{k}

🔷 Step 3 — Use Binomial Expansion

Using:

(1-1)^{18} = \sum_{k=0}^{18} (-1)^k\binom{18}{k} =0

Thus:

\sum_{k=2}^{18} (-1)^k\binom{18}{k} = -\left[ \binom{18}{0} -\binom{18}{1} \right]

=-(1-18)

=17

Therefore:

S= 20\cdot19\cdot17

=6460

🔷 Step 4 — Final Answer

\boxed{6460}

🔷 Step 5 — JEE Trap Alert 🚨

❌ Sign pattern miss kar dena

❌ Combination identity incorrectly apply karna

❌ Binomial sum limits galat lena

Remember:

\sum_{k = 0}^{n} (- 1)^{k} (\binom{n}{k}) = 0

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