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

 

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We want to prove:

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

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Method1: Polynomial Assumption and Coefficient Matching

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

$$\sum_{k=1}^{n} k^2 = An^3 + Bn^2 + Cn + D$$

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

So we assume:

$$\sum_{k=1}^{n} k^2 = An^3 + Bn^2 + Cn$$

Now compute the sum for small values of $n$:

• $n = 1$:

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

• $n = 2$:

  $$1^2 + 2^2 = 1 + 4 = 5 \Rightarrow 8A + 4B + 2C = 5 \tag{2}$$

• $n = 3$:

  $$1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 \Rightarrow 27A + 9B + 3C = 14 \tag{3}$$

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Solving the system of equations

From (1):

$$A + B + C = 1$$

From (2):

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

From (3):

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

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Step 1: Eliminate C

Multiply (1) by 2:

$$2A + 2B + 2C = 2$$

Subtract from (2):

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

Multiply (1) by 3:

$$3A + 3B + 3C = 3$$

Subtract from (3):

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

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

From (4):

$$6A + 2B = 3 \Rightarrow 3A + B = 1.5 \tag{6}$$

From (5):

$$24A + 6B = 11 \Rightarrow 4A + B = \frac{11}{6} \tag{7}$$

Subtract (6) from (7):

$$(4A + B) - (3A + B) = \frac{11}{6} - \frac{3}{2} \Rightarrow A = \frac{11}{6} - \frac{9}{6} = \frac{2}{6} = \frac{1}{3}$$

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

$$3 \cdot \frac{1}{3} + B = 1.5 \Rightarrow 1 + B = 1.5 \Rightarrow B = 0.5 = \frac{1}{2}$$

Now substitute $A$ and $B$ into (1):

$$\frac{1}{3} + \frac{1}{2} + C = 1 \Rightarrow \frac{5}{6} + C = 1 \Rightarrow C = \frac{1}{6}$$

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Final Formula

$$\sum_{k=1}^{n} k^2 = An^3 + Bn^2 + Cn = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n$$

Factor:

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

✅ Q.E.D.

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Method 2: Derivation Using Cubes

We start with the identity:

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

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

Left-hand side (LHS):

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

This is a telescoping sum:

$$(2^3 - 1^3) + (3^3 - 2^3) + (4^3 - 3^3) + \dots + ((N+1)^3 - N^3)$$

Almost everything cancels, and we're left with:

$$(N+1)^3 - 1^3 = (N+1)^3 - 1$$

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Right-hand side (RHS):

$$\sum_{n=1}^N (3n^2 + 3n + 1) = 3 \sum_{n=1}^N n^2 + 3 \sum_{n=1}^N n + \sum_{n=1}^N 1$$

We use known formulas:

• $\sum_{n=1}^N n = \frac{N(N+1)}{2}$

• $\sum_{n=1}^N 1 = N$

So,

$$(N+1)^3 - 1 = 3 \sum_{n=1}^N n^2 + 3 \cdot \frac{N(N+1)}{2} + N$$

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✅ Now Solve for $\sum_{n=1}^N n^2$:

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

Move other terms to the left:

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

Now simplify the RHS:

• $(N+1)^3 = N^3 + 3N^2 + 3N + 1$

• So $(N+1)^3 - 1 = N^3 + 3N^2 + 3N$

Thus:

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

Combine like terms:

$$= N^3 + 3N^2 + 3N - N - \frac{3N(N+1)}{2} = N^3 + 3N^2 + 2N - \frac{3N(N+1)}{2}$$

Now write all with a common denominator:

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

So,

$$3 \sum_{n=1}^N n^2 = \frac{2N^3 + 3N^2 + N}{2}$$

Divide both sides by 3:

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

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✅ Final Result:

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

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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.