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4 years ago

Can someone please help me with this worksheet?

Mathematics

2 answers:

goldenfox [79]4 years ago

8 0

Omg, are you from Slocum?

Aleksandr-060686 [28]4 years ago

5 0

5p-8p=14+4
-3P=18
P=18/3
P=6

X+4X=17-2
5X=15
X=3

-6=3R
2=R

75= -18n-15
90=-18n
5=N

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

Molodets [167]

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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kifflom [539]

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3 years ago

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Anuta_ua [19.1K]

Answer:

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Step-by-step explanation:

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3 years ago

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4 years ago