Question

Find the formula of the series:1³+2³+3³+...+n³​

Answer 1

Explanation:

(x+1)^4=x^4+4x^3+6x^2+4x+1

(X+1)^4–x^4=4x^3+6x^2+4x+1

If x=1>>2^4–1^4=4*1^3+6*1^2+4*1+1

x=2>>3^4–2^4=4*2^3+6*2^2+4*2+1

x=3>>4^4–3^4=4*3^3+6*3^2+6*3+1

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X=n>>(n+1)^4-n^4=4n^3+6n^2+4n+1

X=n-1>>n^4-(n-1)^4=4(n-1)^3+6(n-1)^2+4(n-1)+1

Now we add them together

(n+1)^4–1=4(1^3+2^3+3^3+…+n^3)+6(1^2+2^2+3^2+…+n^2)+4(1+2+3+…+n)+(1+1++1+…+n)

(n+1)^4–1=4s(n)+6{n(n+1)(2n+1)/6}+4{n(n+1)/2}+n . So we could multiply this equation by 6

6(n^4+4n^3+6n^2+4n)=24s(n)+6(2n^3+3n^2+n)+12((n^2+n)+6n

24s(n)=6n^4+24n^3+36n^2+24n-12n^3–18n^2–6n-12n^2–12n-6n

24s(n)=6n^4+12n^3+6n^2

S(n)=(n^4+2n^2+1)/4

answer

(n(n+1)/2)^2

Answer 2

(n²(n+1)²)/4

Explanation:

1³ = 1

1³ + 2³ = 9

1³ + 2³ + 3³ = 36

1³ + 2³ + 3³ + 4³ = 100

Every answer is a perfect square:

1² = 1

3² = 9

6² = 36

10² = 100

Every number to be squared has a similarity:

(1)² = 1² = 1

(1 + 2)² = 3² = 9

(1 + 2 + 3)² = 6² = 36

(1 + 2 + 3 + 4)² = 10² = 100

Therefore we only need to find the formula for 1 + 2 + 3 + ... + n, and square that formula.

For the new series we have:

1 = 1

1 + 2 = 3

1 + 2 + 3 = 6

1 + 2 + 3 + 4 = 10

Let S be the sum of the new series:

S = 1 + 2 + 3 + ... + (n - 1) + n

S can also be re-written as:

S = n + (n - 1) + (n - 2) ... + 2 + 1

Adding the two sums of the series we have:

2S = (n + 1) + (n + 1) + (n + 1) + ... + (n + 1) = n(n + 1)

S = n(n + 1)/2

Now we only need to square this formula to get the formula for the original series:

[n(n + 1)/2]²

Applying Laws of Indices we have:

(n²(n + 1)²)/4

Hope this helps!