What are the solutions to the equation n^2=(n(n+1)(2n+1))/6 ?

The solutions to the equation n^2=(n(n+1)(2n+1))/6 are n=0,n= 1/2 ,n=1

Find the zeros of n^2=(n(n+1)(2n+1))/6

The zeros of n^2=(n(n+1)(2n+1))/6 are n=0,n= 1/2 ,n=1

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

n = 0, n = \frac{1}{2}, n = 1

Decimal Notation

n = 0, n = 0.5, n = 1

Solution steps

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

Multiply both sides by

6

6 n^{2} = n (n + 1) (2 n + 1)

Expand

n (n + 1) (2 n + 1) : 2 n^{3} + 3 n^{2} + n

6 n^{2} = 2 n^{3} + 3 n^{2} + n

Switch sides

2 n^{3} + 3 n^{2} + n = 6 n^{2}

Move

6 n^{2}

to the left side

2 n^{3} - 3 n^{2} + n = 0

Find one solution for

2 n^{3} - 3 n^{2} + n = 0

using Newton-Raphson:

n \approx 0.00000 \dots

Apply long division:

\frac{2 n ^{3} - 3 n ^{2} + n}{n} = 2 n^{2} - 3 n + 1

2 n^{2} - 3 n + 1 \approx 0

Find one solution for

2 n^{2} - 3 n + 1 = 0

using Newton-Raphson:

n \approx 0.50000 \dots

Apply long division:

\frac{2 n ^{2} - 3 n + 1}{n - 0.5} = 2 n - 2

2 n - 2 \approx 0

n \approx 1

Approximate to algebraic form

n = 1

Approximate to algebraic form

n \approx 0.00000 \dots : n = 0

n \approx 0.50000 \dots : n = \frac{1}{2}

n = 0, n = \frac{1}{2}, n = 1

The solutions are

n = 0, n = \frac{1}{2}, n = 1

x^{4} - 34 x^{2} + 255 = 0

4 x^{3} + 2 x - 6 = 0

67 = a^{3}

x^{4} + 3 x - 2 = 0

x^{4} - 3 x^{3} + 2 x^{2} - 3 x + 1 = 0

What are the solutions to the equation n^2=(n(n+1)(2n+1))/6 ?
The solutions to the equation n^2=(n(n+1)(2n+1))/6 are n=0,n= 1/2 ,n=1
Find the zeros of n^2=(n(n+1)(2n+1))/6
The zeros of n^2=(n(n+1)(2n+1))/6 are n=0,n= 1/2 ,n=1