Proof for Sum of squares of first n natural,even,odd numbers - pending

 

We want to prove:

∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

Method1: Polynomial Assumption and Coefficient Matching

The idea is to assume that the sum of squares is a cubic polynomial in nn:

∑k=1nk2=An3+Bn2+Cn+D\sum_{k=1}^{n} k^2 = An^3 + Bn^2 + Cn + D

But since ∑k=10k2=0\sum_{k=1}^0 k^2 = 0, we know that when n=0n = 0, the sum is 0 ⇒ D=0D = 0.

So we assume:

∑k=1nk2=An3+Bn2+Cn\sum_{k=1}^{n} k^2 = An^3 + Bn^2 + Cn

Now compute the sum for small values of nn:

  • n=1n = 1:

    12=A(1)3+B(1)2+C(1)=A+B+C⇒A+B+C=1(1)1^2 = A(1)^3 + B(1)^2 + C(1) = A + B + C \Rightarrow A + B + C = 1 \tag{1}

  • n=2n = 2:

    12+22=1+4=5⇒8A+4B+2C=5(2)1^2 + 2^2 = 1 + 4 = 5 \Rightarrow 8A + 4B + 2C = 5 \tag{2}

  • n=3n = 3:

    12+22+32=1+4+9=14⇒27A+9B+3C=14(3)1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 \Rightarrow 27A + 9B + 3C = 14 \tag{3}

Solving the system of equations

From (1):

A+B+C=1A + B + C = 1

From (2):

8A+4B+2C=58A + 4B + 2C = 5

From (3):

27A+9B+3C=1427A + 9B + 3C = 14

Step 1: Eliminate C

Multiply (1) by 2:

2A+2B+2C=22A + 2B + 2C = 2

Subtract from (2):

(8A+4B+2C)−(2A+2B+2C)=5−2⇒6A+2B=3(4)(8A + 4B + 2C) - (2A + 2B + 2C) = 5 - 2 \Rightarrow 6A + 2B = 3 \tag{4}

Multiply (1) by 3:

3A+3B+3C=33A + 3B + 3C = 3

Subtract from (3):

(27A+9B+3C)−(3A+3B+3C)=14−3⇒24A+6B=11(5)(27A + 9B + 3C) - (3A + 3B + 3C) = 14 - 3 \Rightarrow 24A + 6B = 11 \tag{5}

Step 2: Solve equations (4) and (5)

From (4):

6A+2B=3⇒3A+B=1.5(6)6A + 2B = 3 \Rightarrow 3A + B = 1.5 \tag{6}

From (5):

24A+6B=11⇒4A+B=116(7)24A + 6B = 11 \Rightarrow 4A + B = \frac{11}{6} \tag{7}

Subtract (6) from (7):

(4A+B)−(3A+B)=116−32⇒A=116−96=26=13(4A + B) - (3A + B) = \frac{11}{6} - \frac{3}{2} \Rightarrow A = \frac{11}{6} - \frac{9}{6} = \frac{2}{6} = \frac{1}{3}

Now substitute A=13A = \frac{1}{3} into (6):

3⋅13+B=1.5⇒1+B=1.5⇒B=0.5=123 \cdot \frac{1}{3} + B = 1.5 \Rightarrow 1 + B = 1.5 \Rightarrow B = 0.5 = \frac{1}{2}

Now substitute AA and BB into (1):

13+12+C=1⇒56+C=1⇒C=16\frac{1}{3} + \frac{1}{2} + C = 1 \Rightarrow \frac{5}{6} + C = 1 \Rightarrow C = \frac{1}{6}

Final Formula

∑k=1nk2=An3+Bn2+Cn=13n3+12n2+16n\sum_{k=1}^{n} k^2 = An^3 + Bn^2 + Cn = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n

Factor:

=n(n+1)(2n+1)6= \frac{n(n+1)(2n+1)}{6}

✅ Q.E.D.


Method 2: Derivation Using Cubes

We start with the identity:

(n+1)3−n3=3n2+3n+1(n+1)^3 - n^3 = 3n^2 + 3n + 1

Now, sum both sides from n=1n = 1 to n=Nn = N:

Left-hand side (LHS):

∑n=1N((n+1)3−n3)\sum_{n=1}^N \left((n+1)^3 - n^3\right)

This is a telescoping sum:

(23−13)+(33−23)+(43−33)+⋯+((N+1)3−N3)(2^3 - 1^3) + (3^3 - 2^3) + (4^3 - 3^3) + \dots + ((N+1)^3 - N^3)

Almost everything cancels, and we're left with:

(N+1)3−13=(N+1)3−1(N+1)^3 - 1^3 = (N+1)^3 - 1

Right-hand side (RHS):

∑n=1N(3n2+3n+1)=3∑n=1Nn2+3∑n=1Nn+∑n=1N1\sum_{n=1}^N (3n^2 + 3n + 1) = 3 \sum_{n=1}^N n^2 + 3 \sum_{n=1}^N n + \sum_{n=1}^N 1

We use known formulas:

  • ∑n=1Nn=N(N+1)2\sum_{n=1}^N n = \frac{N(N+1)}{2}

  • ∑n=1N1=N\sum_{n=1}^N 1 = N

So,

(N+1)3−1=3∑n=1Nn2+3⋅N(N+1)2+N(N+1)^3 - 1 = 3 \sum_{n=1}^N n^2 + 3 \cdot \frac{N(N+1)}{2} + N

✅ Now Solve for ∑n=1Nn2\sum_{n=1}^N n^2:

(N+1)3−1=3∑n=1Nn2+3N(N+1)2+N(N+1)^3 - 1 = 3 \sum_{n=1}^N n^2 + \frac{3N(N+1)}{2} + N

Move other terms to the left:

3∑n=1Nn2=(N+1)3−1−3N(N+1)2−N3 \sum_{n=1}^N n^2 = (N+1)^3 - 1 - \frac{3N(N+1)}{2} - N

Now simplify the RHS:

  • (N+1)3=N3+3N2+3N+1(N+1)^3 = N^3 + 3N^2 + 3N + 1

  • So (N+1)3−1=N3+3N2+3N(N+1)^3 - 1 = N^3 + 3N^2 + 3N

Thus:

3∑n=1Nn2=N3+3N2+3N−3N(N+1)2−N3 \sum_{n=1}^N n^2 = N^3 + 3N^2 + 3N - \frac{3N(N+1)}{2} - N

Combine like terms:

=N3+3N2+3N−N−3N(N+1)2=N3+3N2+2N−3N(N+1)2= N^3 + 3N^2 + 3N - N - \frac{3N(N+1)}{2} = N^3 + 3N^2 + 2N - \frac{3N(N+1)}{2}

Now write all with a common denominator:

=2N3+6N2+4N−3N(N+1)2=2N3+6N2+4N−3N2−3N2=2N3+3N2+N2= \frac{2N^3 + 6N^2 + 4N - 3N(N+1)}{2} = \frac{2N^3 + 6N^2 + 4N - 3N^2 - 3N}{2} = \frac{2N^3 + 3N^2 + N}{2}

So,

3∑n=1Nn2=2N3+3N2+N23 \sum_{n=1}^N n^2 = \frac{2N^3 + 3N^2 + N}{2}

Divide both sides by 3:

∑n=1Nn2=2N3+3N2+N6=N(N+1)(2N+1)6\sum_{n=1}^N n^2 = \frac{2N^3 + 3N^2 + N}{6} = \frac{N(N+1)(2N+1)}{6}

✅ Final Result:

∑k=1nk2=n(n+1)(2n+1)6\boxed{\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}}


From here, it's easy to prove for even number sum: multiply by 4.
Then subtract that from sum of first 2n numbers' squares to get the formula for odd number square sum.

Comments

Post a Comment

Popular posts from this blog

IOQM 2024 Paper solutions (Done 1-21, 29)

Image
Q1 (number-theory). The smallest positive integer that does not divide 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 is: Solution : This product has all the integers from 1 to 9. So smallest integer not dividing it would be the next prime after 9. Which is 11. Answer: 11 Q2 (combinatorics): The number of four-digit odd numbers having digits 1, 2, 3, 4, each occurring exactly once, is: Solution: Last digit has to be 1 or 3: 2C1 Choosing remaining 3 digits: 3C3 Arranging those 3 digits: 3! Answer: 2C1*3C3*3! = 12 Q3 (number-theory): The number obtained by taking the last two digits of \(5^{2024}\) in the same order is: Solution: To get last 2 digits from number, divide by 100 and take the remainder. So we have to compute mod 100. 25 mod 100 = 25 25.5 mod 100 = 125 mod 100 = 25 5^3.5 mod 100 = 25.5 mod 100 = 25 5^4.5 mod 100 = 25.5 mod 100 = 25 ..... So: 5^2024 mod 100 = 25. Q4 (geometry): Let ABCD be a quadrilateral with \(\angle ADC = 70^\circ\), \(\angle ACD = 70^\circ\), \(\angle ACB = 10^\circ\...
Read more

Combinatorics DPP - RACE 6 - Q16 pending discussion

Image
Q1 . A question paper consists of two sections. Section A has 7 questions and section B has 8 questions. A student has to answer 10 questions out of these 15 questions. The number of ways in which he can answer if he must answer at least 3 of section A and at least 4 of the section B is: Solution: One wrong way to solve this: For the mandatory questions: 7C3*8C4 Now 3 more questions need to be answered from 4 of A and 4 of B. Cases: 3A,0B | 2A,1B | 1A,2B | 0A,3B 4C3*4C0 + 4C2*4C1 + 4C1*4C2 + 4C0*4C1 = 4 + 24 + 24 + 4 = 56 Finally: 7C3*8C4*56 = 35*70*56 = 137200 Why is it wrong? Let's say we chose A1,A2,A3 as mandatory from A and then chose A4,A5,A6 when we did 4C3*4C0 for (3A,0B). But we could have done it the other way round also, i.e. choose A4,A5,A6 as mandatory and then choose A1,A2,A3. So there is overcounting. Correct Solution : Simply create the cases for A,B: (3,7),(4,6),(5,5),(6,4) 7C3*8C7 + 7C4*8C6 + 7C5*8C5 + 7C6*8C4 = 2926. Q2 . A kindergarten teacher has 25 kids in her...
Read more

Algebra DPP 1.3 Quadratics

 Q1 . For what values of b do the equations: 1988x² + bx + 8891 = 0 and 8891x² + bx + 1988 = 0 have a common root? Solution: We are given two quadratic equations: 1988 x 2 + b x + 8891 = 0 1988x^2 + bx + 8891 = 0 8891 x 2 + b x + 1988 = 0 8891x^2 + bx + 1988 = 0 We are to find the values of b b for which they have a common root . Step 1: Let α \alpha be the common root. Then α \alpha satisfies both equations. So, From equation (1): 1988 α 2 + b α + 8891 = 0 (i) 1988\alpha^2 + b\alpha + 8891 = 0 \quad \text{(i)} From equation (2): 8891 α 2 + b α + 1988 = 0 (ii) 8891\alpha^2 + b\alpha + 1988 = 0 \quad \text{(ii)} Step 2: Subtract equation (i) from equation (ii): ( 8891 α 2 − 1988 α 2 ) + ( b α − b α ) + ( 1988 − 8891 ) = 0 (8891\alpha^2 - 1988\alpha^2) + (b\alpha - b\alpha) + (1988 - 8891) = 0 ( 8891 − 1988 ) α 2 − ( 8891 − 1988 ) = 0 (8891 - 1988)\alpha^2 - (8891 - 1988) = 0 6903 α 2 − 6903 = 0 6903\alpha^2 - 6903 = 0 α 2 = 1 ⇒ α = ± 1 \alpha^2 = 1 \Rightar...
Read more