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2 years ago

5

Priya is stacking building blocks to make a tower. She takes a break when the tower is 212 feet tall, which is 58 of the height

of the tower she wants to build. How tall is the tower when finished?

1 answer:

aleksklad [387]2 years ago

7 0

Answer:

4 feet.

Step-by-step explanation:

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I really stuck on trying to prove this

Alisiya [41]

$\displaystyle\sum_{r=1}^nr(r+1)\cdots(r+p-1)$

When $n=1$ ,

$\displaystyle\sum_{r=1}^1r(r+1)\cdots(r+p-1)=1(1+1)(1+2)\cdots(1+p-2)(1+p-1)=p!$

Meanwhile, you have on the right

$\dfrac{(1)(1+1)(1+2)\cdots(1+p-2)(1+p-1)(1+p)}{p+1}=(1)(1+1)(1+2)\cdots(p-1)(p)=p!$

so the equality holds for $n=1$ .

Assume it holds for $n=k$ , i.e. that

$\displaystyle\sum_{r=1}^kr(r+1)\cdots(r+p-1)=\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}$

Now for $n=k+1$ , you have

$\displaystyle\sum_{r=1}^{k+1}r(r+1)\cdots(r+p-1)=\sum_{r=1}^kr(r+1)\cdots(r+p-1)+(k+1)(k+2)\cdots(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+1+p-2)(k+1+p-1)$
$=\displaystyle\frac{k(k+1)(k+2)\cdots(k+p-1)(k+p)}{p+1}+(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\left(\dfrac k{p+1}+1\right)(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{k+p+1}{p+1}(k+1)(k+2)\cdots(k+p-1)(k+p)$
$=\dfrac{(k+1)(k+2)\cdots(k+p-1)(k+p)(k+p+1)}{p+1}$

as required.

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<span>d = sqrt ((2 - -2)^2 + (-7 - 3)^2 +(4 - -5)^2) = sqrt 197
</span>
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