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# Diamond Pattern | AMC-10A, 2009 | Problem 15

Try this beautiful problem from AMC 10A, 2009 based on Diamond Pattern.

## Diamond Pattern - AMC-10A, 2009- Problem 15

The figures $F_1$, $F_2$, $F_3$, and $F_4$ shown are the first in a sequence of figures. For $n\ge3$, $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$?

• $756$
• $761$
• $786$

### Key Concepts

Pattern

Sequence

Symmetry

Answer: $761$

AMC-10A (2009) Problem 15

Pre College Mathematics

## Try with Hints

From the above diagram we observe that in $F_1$ the number of diamond is $1$.in $F_2$ the number of diamonds are $5$.in $F_3$ the number of diamonds are $13$. in $F_4$ the numbers of diamonds are $25$.Therefore from $F_1$ to $F_2$ ,$(5-1)$=$4$ new diamonds added.from $F_2$ to $F_3$,$(13-5=8$ new diamonds added.from $F_3$ to $F_4$,$(25-13)=12$ new diamonds added.we may say that When constructing $F_n$ from $F_{n-1}$, we add $4(n-1)$ new diamonds.

Can you now finish the problem ..........

so we may construct that Let $S_n$ be the number of diamonds in $F_n$. We already know that $P_1$=1 and for all $n >1$ ,$P_n=P_{n-1}+4(n-1)$.now can you find out $P_{20}$

can you finish the problem........

Now $P_{20}$=$P_{19} + 4(20-1)$

$\Rightarrow P_{20}$=$P_{19} + 4.19)$

$\Rightarrow P_{20}$=$P_{18}+(4 \times 18) +( 4 \times 19)$

$\Rightarrow P_{20}$= .........................................

$\Rightarrow P_{20}$=$1+4(1+2+3+........+18+19)$

$\Rightarrow P_{20}$=$1+ \frac{ 4 \times 19 \times 20}{2}$

$\Rightarrow P_{20}$=$761$