JEE Algebra Focus Reading Time: 12 min 3 Key Formulas

Summation of Special Series: ∑n, ∑n², ∑n³ - Complete Guide

Master the fundamental summation formulas with geometric proofs, derivations, and JEE-level applications.

Key Formulas

100%

JEE Relevance

Proof Methods

15min

Practice Time

Why These Summation Formulas Are Essential

The summation formulas for ∑n, ∑n², and ∑n³ form the foundation of series and sequences in JEE Mathematics. These appear in:

Direct formula application problems (JEE Main)
Proof-based questions (JEE Advanced)
Method of differences applications
Integration approximation problems

Formula 1 Essential

Sum of First n Natural Numbers

$$S_1 = \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$$

📐 Geometric Proof (Gauss Method):

Step 1: Write the sum forward and backward:

$S = 1 + 2 + 3 + \cdots + (n-1) + n$

$S = n + (n-1) + (n-2) + \cdots + 2 + 1$

Step 2: Add both equations:

$2S = (n+1) + (n+1) + (n+1) + \cdots + (n+1)$

Step 3: There are n terms of (n+1):

$2S = n(n+1)$

Step 4: Final formula:

$S = \frac{n(n+1)}{2}$

🎯 JEE Application Example:

Problem: Find the sum of first 100 natural numbers.

Solution: Using the formula:

$\sum_{k=1}^{100} k = \frac{100 \times 101}{2} = 50 \times 101 = 5050$

Formula 2 Important

Sum of Squares of First n Natural Numbers

$$S_2 = \sum_{k=1}^n k^2 = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$

📐 Proof Using Telescoping Series:

Step 1: Use the identity: $(k+1)^3 - k^3 = 3k^2 + 3k + 1$

Step 2: Write for k = 1 to n:

$2^3 - 1^3 = 3\cdot1^2 + 3\cdot1 + 1$

$3^3 - 2^3 = 3\cdot2^2 + 3\cdot2 + 1$

$\vdots$

$(n+1)^3 - n^3 = 3\cdot n^2 + 3\cdot n + 1$

Step 3: Add all equations (LHS telescopes):

$(n+1)^3 - 1 = 3\sum k^2 + 3\sum k + n$

Step 4: Substitute $\sum k = \frac{n(n+1)}{2}$:

$n^3 + 3n^2 + 3n = 3\sum k^2 + \frac{3n(n+1)}{2} + n$

Step 5: Solve for $\sum k^2$:

$\sum k^2 = \frac{n(n+1)(2n+1)}{6}$

🎯 JEE Application Example:

Problem: Find $1^2 + 2^2 + 3^2 + \cdots + 20^2$

Solution: Using the formula:

$\sum_{k=1}^{20} k^2 = \frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870$

Formula 3 Advanced

Sum of Cubes of First n Natural Numbers

$$S_3 = \sum_{k=1}^n k^3 = 1^3 + 2^3 + 3^3 + \cdots + n^3 = \left[\frac{n(n+1)}{2}\right]^2$$

📐 Proof Using Mathematical Induction:

Step 1: Base case (n=1):

LHS = $1^3 = 1$, RHS = $\left[\frac{1\times2}{2}\right]^2 = 1$ ✓

Step 2: Assume true for n = m:

$1^3 + 2^3 + \cdots + m^3 = \left[\frac{m(m+1)}{2}\right]^2$

Step 3: Prove for n = m+1:

$1^3 + 2^3 + \cdots + m^3 + (m+1)^3 = \left[\frac{m(m+1)}{2}\right]^2 + (m+1)^3$

$= (m+1)^2\left[\frac{m^2}{4} + (m+1)\right]$

$= (m+1)^2\left[\frac{m^2 + 4m + 4}{4}\right]$

$= \left[\frac{(m+1)(m+2)}{2}\right]^2$ ✓

🎯 JEE Application Example:

Problem: Find $1^3 + 2^3 + 3^3 + \cdots + 10^3$

Solution: Using the formula:

$\sum_{k=1}^{10} k^3 = \left[\frac{10 \times 11}{2}\right]^2 = [55]^2 = 3025$

Note: This equals $(1+2+3+\cdots+10)^2$

🚀 Problem-Solving Strategies

Memory Techniques:

∑n³ = (∑n)² - Beautiful pattern!
∑n² denominator is always 6
All formulas have n(n+1) factor
For ∑n²: multiply by (2n+1)

JEE Exam Tips:

Verify formulas for n=1,2,3
Use method of differences for variations
Remember special cases for quick answers
Practice mixed series problems

Advanced Applications Available

Includes method of differences, summation of special series, and JEE Advanced level problems

📝 Quick Self-Test

Try these JEE-level problems to test your understanding:

1. Find $\sum_{k=1}^{15} (2k-1)$ [Sum of first 15 odd numbers]

2. Calculate $1^2 + 3^2 + 5^2 + \cdots + 19^2$

3. Prove that $1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2$

Ready to Master Series & Sequences?

Get complete access to advanced summation techniques and JEE practice problems