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baherus [9]
3 years ago
8

Evaluate 1^3 + 2^3 +3^3 +.......+ n^3

Mathematics
1 answer:
Molodets [167]3 years ago
7 0

Notice that

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

so that

\displaystyle\sum_{i=1}^n((n+1)^4-n^4)=\sum_{i=1}^n(4i^3+6i^2+4i+1)

We have

\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(2^4-1^4)+(3^4-2^4)+(4^4-3^4)+\cdots+((n+1)^4-n^4)

\implies\displaystyle\sum_{i=1}^n((i+1)^4-i^4)=(n+1)^4-1

so that

\displaystyle(n+1)^4-1=\sum_{i=1}^n(4i^3+6i^2+4i+1)

You might already know that

\displaystyle\sum_{i=1}^n1=n

\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2

\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6

so from these formulas we get

\displaystyle(n+1)^4-1=4\sum_{i=1}^ni^3+n(n+1)(2n+1)+2n(n+1)+n

\implies\displaystyle\sum_{i=1}^ni^3=\frac{(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n}4

\implies\boxed{\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4}

If you don't know the formulas mentioned above:

  • The first one should be obvious; if you add n copies of 1 together, you end up with n.
  • The second one is easily derived: If S=1+2+3+\cdots+n, then S=n+(n-1)+(n-2)+\cdots+1, so that 2S=n(n+1) or S=\dfrac{n(n+1)}2.
  • The third can be derived using a similar strategy to the one used here. Consider the expression (n+1)^3-n^3=3n^2+3n+1, and so on.
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Step-by-step explanation:

This factoring job lends itself well to synthetic division.  Looking at the constant term, -4, I came up with several possible roots based upon -4:  {±1, ±2, ±4}.  I chose +4 as my first trial root.  Sure enough, there was a zero remainder, which indicated that 4 is a root of this polynomial and (x - 4) is a factor.  The coefficients of the trinomial quotients are 1   0   1, which indicates a quotient of x^2 + 1, which has the following roots:  x = +(i) and x = -(i)

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