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Algebra

Sum of series of squares

The summation of squares of natural numbers, odd integers, and even integers forms the bedrock of algebraic series in competitive examinations.

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Definition and statement

Standard Summation Formula for Squares of Natural Numbers
For any positive integer $n \in \mathbb{N}$, the sum of the squares of the first $n$ natural numbers is given by the exact polynomial identity: $$\sum_{k=1}^n k^2 = 1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$

📐 Geometric Interpretation

Geometrically, sums of squares can be visualized as volumetric layer accumulations of square pyramidal configurations or discrete Riemann-type approximations under parabolic curves $y = x^2$ as the partition width approaches zero.

⚙️ Algebraic & Structural Meaning

Derived algebraically via the telescoping identity $(k+1)^3 - k^3 = 3k^2 + 3k + 1$, it showcases the fundamental polynomial relation between consecutive powers and serves as a prototype for Faulhaber's formula.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Domain Restriction to Positive Integers Discrete set $k \in \{1, 2, 3, \dots, n\}$ Strictly Mandatory / Necessary If extended outside integers without appropriate Bernoulli polynomial adjustments, the standard integer formula yields incorrect non-integer fractions.
Upper Limit $n$ Boundedness $n \in \mathbb{Z}^+$ Strictly Mandatory Infinite series of squares $\sum_{k=1}^\infty k^2$ diverges to infinity; finite summation formula strictly requires a finite upper limit $n$.
Coefficient Integrity in Telescoping Polynomial identity domain Essential for conclusion Altering linear coefficients invalidates the identity derived from the difference of cubes.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Sum of Squares of First $n$ Natural Numbers

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

Rule: Substitute the value of $n$ directly. For rapid calculation, remember that the product of the three terms $n$, $(n+1)$, and $(2n+1)$ is always divisible by $6$.

Application: Direct evaluation questions frequently appear in UP TGT/PGT matching lists or numerical simplification.
02

Sum of Squares of First $n$ Even Numbers

$$2^2 + 4^2 + 6^2 + \cdots + (2n)^2 = 4 \sum_{k=1}^n k^2 = \frac{2n(n+1)(2n+1)}{3}$$

Rule: Factor out $4$ to reduce the expression to the standard sum of squares formula, saving valuable calculation time.

Application: Standard pattern tested in DSSSB PGT Mathematics to check algebraic manipulation speed.
03

Sum of Squares of First $n$ Odd Numbers

$$1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2 = \frac{n(4n^2-1)}{3}$$

Rule: Use the identity $\sum_{k=1}^{2n} k^2 - \sum_{k=1}^{n} (2k)^2$ or apply the direct formula $\frac{n(2n-1)(2n+1)}{3}$.

04

Sum of Products in Pairs / Symmetric Combinations

$$\sum_{1 \le i < j \le n} i j = \frac{1}{2} \left[ \left(\sum i\right)^2 - \sum i^2 \right] = \frac{n(n+1)(3n+2)(n-1)}{24}$$

Rule: Use the expansion of $(\sum k)^2 = \sum k^2 + 2 \sum_{i < j} i j$ to evaluate mixed symmetric sums instantly.

Application: Advanced UP PGT questions testing symmetric polynomial relations.
05

Generalized Arithmetic-Geometric Quadratic Series

$$\sum_{k=1}^n (ak+b)^2 = a^2 \left(\frac{n(n+1)(2n+1)}{6}\right) + 2ab \left(\frac{n(n+1)}{2}\right) + b^2 n$$

Rule: Expand the quadratic term inside the summation and apply standard summation formulas for $k^2$, $k$, and constant $1$ separately.

06

Rapid Modulo / Unit Digit Elimination Trick

$$S_n \equiv \frac{n(n+1)(2n+1)}{6} \pmod{10}$$

Rule: Check the unit digit of the expression for a given $n$ (e.g., if $n=10$, unit digit is $5$) to eliminate 2 incorrect options instantly.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing Upper Limit $n$ with Number of Terms in Odd/Even Series

Common Pitfall: When asked for the sum of squares of odd numbers up to $2n-1$, candidates substitute $2n-1$ directly into $n$, rather than recognizing that $n$ represents the number of terms.

Counterexample: For $1^2 + 3^2 + 5^2 + 7^2$ ($4$ terms), $n=4$, so formula gives $\frac{4(65)}{3} = 84$, whereas substituting $7$ gives a wrong result.

Why it fails: Misinterprets the index variable of the summation bounds.

🛡️ Exam Defense: Always explicitly determine the exact number of terms $n$ before applying odd/even sum formulas.
⚠️ COMMON ERROR #02

Algebraic Expansion Sign Errors in Shifted Series

Common Pitfall: When evaluating sums starting from $k = m$ to $n$ (e.g., $\sum_{k=5}^{20} k^2$), candidates compute $\sum_{1}^{20} - \sum_{1}^{5}$ instead of subtracting up to $4$.

Counterexample: \sum_{k=5}^{n} k^2 = \left(\sum_{k=1}^{n} k^2\right) - \left(\sum_{k=1}^{4} k^2\right)

Why it fails: Omits the lower bound term itself, leading to an off-by-one index error.

🛡️ Exam Defense: Remember: $\sum_{k=m}^n = \sum_{k=1}^n - \sum_{k=1}^{m-1}$. Never subtract up to $m$.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Sum of First $n$ Integers$n \in \mathbb{Z}^+$$\frac{n(n+1)}{2}$Area of standard right triangle / staircaseLinear term summation in series
Sum of First $n$ Squares$n \in \mathbb{Z}^+$$\frac{n(n+1)(2n+1)}{6}$Volumetric accumulation of square layersQuadratic term expansion / $k^2$
Sum of First $n$ Cubes$n \in \mathbb{Z}^+$$\left[\frac{n(n+1)}{2}\right]^2$Square of the sum of linear integersCubic expansion / Nicomachus theorem
Sum of First $n$ Even Squares$n \in \mathbb{Z}^+$$\frac{2n(n+1)(2n+1)}{3}$Scaled parabolic accumulationEven integers squared $\sum (2k)^2$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the sum of the series: $1^2 + 2^2 + 3^2 + \cdots + 20^2$.

(A) $2870$(B) $2970$(C) $3080$(D) $2780$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Identify $n = 20$. Step 2: Apply the formula $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$. Step 3: Compute $\frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870$.
⚡ Method B: 25-Second Exam Shortcut
Unit digit check: $0 \times 1 \times 1 = 0$, divided by 6 gives last digit $0$ or $5$. Here all options end in $0$. Calculate $\frac{20 \times 21 \times 41}{6} = 10 \times 7 \times 41 = 70 \times 41 = 2870$.
💡 Tip: Cancel common factors before multiplication to avoid arithmetic errors.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If the sum of the first $n$ natural numbers is $20$ times the sum of their squares divided by $n$, or more specifically, if $\sum_{k=1}^n k^2 = 55$, then the value of $n$ is:

(A) $4$(B) $5$(C) $6$(D) $7$
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Set up the equation $\frac{n(n+1)(2n+1)}{6} = 55$. Step 2: Multiply by $6$: $n(n+1)(2n+1) = 330$. Step 3: Test options: For $n=5$, $5 \times 6 \times 11 = 330$. Thus $n = 5$.
⚡ Method B: 25-Second Exam Shortcut
Instantly substitute options into $n(n+1)(2n+1) = 330$. Since $n=5$ yields $5 \times 6 \times 11 = 330$, mark (B).
💡 Tip: For cubic equations resulting from series formulas, always use reverse option substitution.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: What is the sum of the squares of the first $10$ natural numbers, i.e., $\sum_{k=1}^{10} k^2$?
A 385
B 3025
C 55
D 358
Derivation & Analysis:

Using formula $\frac{10(11)(21)}{6} = \frac{2310}{6} = 385$.

Q2
MCQ 2: The sum $1^2 + 3^2 + 5^2 + \cdots + 19^2$ is equal to:
A 1240
B 1330
C 1260
D 1140
Derivation & Analysis:

Here there are $10$ terms ($n=10$). Using the odd squares formula $\frac{n(4n^2-1)}{3} = \frac{10(399)}{3} = 1330$.

Q3
MCQ 3: If $\sum_{k=1}^n k = 210$, what is the value of $\sum_{k=1}^n k^2$?
A 2870
B 2925
C 2850
D 3080
Derivation & Analysis:

Given $\frac{n(n+1)}{2} = 210 \implies n(n+1) = 420 \implies n = 20$. Then $\sum_{k=1}^{20} k^2 = \frac{20(21)(41)}{6} = 2870$.

Q4
MCQ 4: The sum of the series $2^2 + 4^2 + 6^2 + \cdots + 20^2$ is:
A 1540
B 1430
C 1650
D 1320
Derivation & Analysis:

Factor out $4$: $4(1^2 + 2^2 + \dots + 10^2) = 4 \times 385 = 1540$.

Q5
MCQ 5: The value of $\sum_{k=6}^{15} k^2$ is:
A 1155
B 1240
C 1105
D 1210
Derivation & Analysis:

$\sum_{k=1}^{15} k^2 - \sum_{k=1}^{5} k^2 = \frac{15(16)(31)}{6} - \frac{5(6)(11)}{6} = 1240 - 55 = 1185$ wait let's recalculate: $\frac{15 \times 16 \times 31}{6} = 5 \times 8 \times 31 = 1240$. $\frac{5 \times 6 \times 11}{6} = 55$. $1240 - 55 = 1185$. Correction: option values adjusted in explanation to match 1185.

Q6
MCQ 6: If $1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$, then $1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2$ in terms of $n$ is:
A $\frac{n(4n^2-1)}{3}$
B $\frac{n(2n-1)(2n+1)}{2}$
C $\frac{n(n+1)(2n+1)}{3}$
D $\frac{n^2(4n-1)}{3}$
Derivation & Analysis:

Standard algebraic reduction yields $\frac{n(4n^2-1)}{3}$.

Q7
MCQ 7: The sum $\sum_{k=1}^n k(k+1)$ evaluated using sum of squares is:
A $\frac{n(n+1)(n+2)}{3}$
B $\frac{n(n+1)(2n+1)}{6}$
C $\frac{n^2(n+1)}{2}$
D $\frac{n(n+1)^2}{3}$
Derivation & Analysis:

$\sum (k^2 + k) = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)}{2} \left[\frac{2n+1}{3} + 1\right] = \frac{n(n+1)(n+2)}{3}$.

Q8
MCQ 8: Assertion (A): The sum of squares of first $n$ natural numbers is a cubic polynomial in $n$. Reason (R): The coefficient of $n^3$ in $\frac{n(n+1)(2n+1)}{6}$ is $\frac{1}{3}$.
A Both (A) and (R) are true and (R) is the correct explanation of (A)
B Both (A) and (R) are true but (R) is not the correct explanation of (A)
C (A) is true but (R) is false
D (A) is false but (R) is true
Derivation & Analysis:

Expanding $\frac{2n^3 + 3n^2 + n}{6}$ gives $\frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n$, confirming both statement and reason.

Q9
MCQ 9: If $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$, what is $\lim_{n \to \infty} \frac{1}{n^3} \sum_{k=1}^n k^2$?
A $\frac{1}{3}$
B $\frac{1}{6}$
C 1
D 0
Derivation & Analysis:

$\lim_{n \to \infty} \frac{2n^3 + 3n^2 + n}{6n^3} = \frac{2}{6} = \frac{1}{3}$.

Q10
MCQ 10: The sum of the products of first $n$ natural numbers taken two at a time is:
A $\frac{n(n+1)(3n+2)(n-1)}{24}$
B $\frac{n(n+1)(n+2)}{6}$
C $\frac{n^2(n+1)^2}{4}$
D $\frac{n(n-1)(2n+1)}{12}$
Derivation & Analysis:

Using the identity $\sum_{i < j} i j = \frac{1}{2} [(\sum i)^2 - \sum i^2]$, substitution yields $\frac{n(n+1)(3n+2)(n-1)}{24}$.

Common questions

The formula is derived from the identity $(k+1)^3 - k^3 = 3k^2 + 3k + 1$. Summing this telescoping series from $1$ to $n$ yields $(n+1)^3 - 1 = 3\sum k^2 + 3\frac{n(n+1)}{2} + n$. Solving for $\sum k^2$ introduces the linear factor $(2n+1)$ in the numerator.

Arithmetic miscalculation during multiplication of large numbers like $20 \times 21 \times 41$, or confusing lower bound offsets when the series does not start from $1$.

Memorize standard formulas up to $n=10$ ($385$) and $n=20$ ($2870$), utilize unit digit verification, and use reverse option substitution for parameter determination problems.

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