Question

mili used cube blocks to build some solids. according to this pattern, how many cube blocks do we need to prepare to build the 9th solid? 1st 2nd 3rd

Explanation:

Step1: Analyze the pattern

Let's find the number of blocks in each solid. The first solid has $1\times5 + 1\times4=9$ blocks. The second solid has $2\times5+2\times4 + 1\times3=21$ blocks. The third solid has $3\times5+3\times4+2\times3 = 33$ blocks.
We can observe that the number of blocks $N$ in the $n^{th}$ solid can be calculated by the formula $N = n\times5 + n\times4+(n - 1)\times3$. Simplify this formula:
\[

$$\begin{align*} N&=5n+4n + 3n-3\\ &=(5 + 4+3)n-3\\ &=12n-3 \end{align*}$$

\]

Step2: Calculate for the 9th solid

Substitute $n = 9$ into the formula $N = 12n-3$.
$N=12\times9-3$
$N = 108 - 3$
$N=105$

Answer:

105