Area Patterns!

$\begin{aligned} &1& = 1 \\ &5& = 1+4 \\ &9& = 1+4+4 \\ &13& = 1+4+4+4 \\ &\boxed{17}& = 1+4+4+4+4 \end{aligned}$

In every figure, the number of square boxes increase by 4. So, likewise in figure 5 there are 17 square boxes

a=1; d=4, 5th term T5=?

Tn=a+(n-1) d, T5=1+(5-1) 4, T5=17, Total 17 squires are present in that.

It follows an Arithmetic Progression with a common difference(d) of 4.

so, $T_{5}$ = $1+(5-1)*4 = 17$

thus, its an esy AP.

cheers!!

It works like and A.P. with first term as 1 and common difference of 4..

the number of squares +4

The rule is n2+1

$S_n=a(n-1)+1$

where:

$n=1,2,3....., \infty$

$a=4$

for $n=5$ , $\boxed{S_n=17}$

it is very simple , just add 4 units to the last figure = 13 + 4 = 17

Okay , it is good =) , I solved it this way ( 4( the figure number -1) + 1 ) , let's try it , the second figure will be (4(1)+1=5) , so the fifth one is (4(4)+1=17) ..... I hope it's understood ..

Number of square keeps on increasing by 4 in every figure.

Every figure has 4 more squares than the figure before it. In other words, 1+[( $n$ -1)×4] where $n$ is the figure number. So, if $n$ is 5, then 1+[(5-1)×4] or 16+1, which is 17.

Each step adds 4 squares. This is a linear progression and the result is obvious.

1=1, 4+1=5, 4+4+1=9, 4+4+4+1=13, 4+4+4+4+1=17. I must say that in every square block the no. is increased by 4 i.e the answer is 17

We can solve it by finding the expression which can help us to solve for the nth term

The expression of the above-mentioned pattern is:

$=4n-3$

And if we input 5, it becomes:

$=(4*5)-3$

it is equal to:

$\boxed{=17}$

In every figure 4 boxes are adding up. Therefore In figure 4 there are 13 boxes. So, if we continue the figure then in the 5th figure it will be 13+4=17

very simple

Add 4 more squares to every figure.

y = 1 + (x-1)4

fig 1 = 1 + 0 = 1

fig 2 = 1 + (1)4 = 5

fig 3 = 1 + (3-1)4 = 9

fig 4 = 1 + (4-1)4 = 13

fig 5 = 1 + (5-1)4 = 1 + 4(4) = 1 + 16 = 17

Good one, and needed.

Add4after eachdiagram

Let n be the number of squares in one of the "square lines", not counting the middle one. Then there is the equation 4n + 1. Nothing wrong with the difference of 4, just that with big numbers its easier to use equations.

It's a simple progression, wich can be expressed by the formulæ $1 + 4(n - 1)$ . For the question asks for the $5^{th}$ number, it's easy to solve, but if you want to know a larger number, using the formulæ is easier.

Application of arithmetic sequence in finding the nth term. an=a1+(n-1)d where: a1-the first term (figure 1) n-number of terms which is 5 and, d-common difference which is 4 Substitute all the values and there you have it! The answer is 17 :)

Add 4 squares to every fig. u'll get answer fig1 = 1 square fig 2 = 1+4 squares=5squares fig3 = 5+4squares=9squares fig4 = 9+squares=13squares fig5 = 13+4squares=17squares(answer)

If we see each figure as an element of a sequence, we can see that the total number of squares present in each figure can be modelled by this function: $m(n) = (n - 1) \times 4 + 1$ .

Because we want the number of squares present in figure 5, we let $n = 5$ , and, using the model, we can see that there must be 17 squares present in figure 5.

first grade math, count the figure 4 and add 4 to get 17...

Plus four to preceding number

Just add 4 to fig.4

Each additional figure adds on four. Fig. 1 = 1 | Fig. 2 = 5 | Fig 3 ... and so on

Every time 4sqares are added.

a=1; d=4, 5th term T5=? Tn=a+(n-1)d, T5=1+(5-1)4, T5=17, Total 17 squires are present in that.

Follow the pattern one is added to every side each time

Add 4 to the 4th item

First 1=1 Second 1+4=5 Third 1+4+4=9 Forth 1+4+4+4=13 Fifth 1+4+4+4+4=17

Using Arithmetic Progression 1,5,9,13 Difference =4

So ,13+4=17

1+4=5+4=9+=13+4=17

4x + 1 where x=4

Each sequence just adds 4

1, 5, 9, 13, 17

1+(5-1)4=17

simply an Arithmetic Progression where $a_1=1$ and common difference $d=4$ ... so for Figure 5, where n = 5, the number of squares will be... $a_5=1+4\times4=17~squares$

We notice that 1st fig shows 1 square and the 2nd fig shows fig1 plus 4 squares After fig3 shows fig2 plus 4 squares.If we want to have a general equation for area of cross, it will be for a figure n An=1+(n-1)4,for n>=1 As a result, for n=5 we will take A_5=1+4*4=1+16=17

the number of squares in the $n^{th}$ square can be written as : 4n-3

The general formula for such a problem is 4n+1. In Fig. 1, its 4(0)+1=1; in Fig. 2, its 4(1)+1=5; in Fig. 3, its 4(2)+1=9; in Fig. 4, its 4(3)+1=13; Thus, in Fig. 5 (not shown), it should be 4(4)+1=17.

By simply looking we can notice that if we hide the center part we can have a pattern of 4(n-1) so by adding 1, the center, we can get the pattern expression of 4(n-1)+1 and by substitution we can get 17.

rule X = 1 + n*4

1 + (4 X 4) =17 where 4 are added at each stage and there are 4 such addition stages

in simple words the center square will be surrounded by n-1 sqaures ...where 'n' is the corresponding number of the figures..

My formula is x=(n * 2) -1

Examples: Fig.1: 1 * 2 = 2 - 1 = 1 Fig.2: 3 * 2 = 6 - 1 = 5 Fig.3: 5 * 2 =10 -1= 9 Fig.4: 7 * 2 =14 -1= 13 Fig.5: 9 * 2 =18 -1= 17

If the past figure contains 'n' squares,the next figure will contain 'n+4' squares

The formula to calculate the area of the $nth$ figure is $4n-3$ because we added 4 each time. We just plug in 5 and get $\boxed{17}$ as our answer.

every time it get (+4)

1+4 5+4 9+4 13+4 simple pattern if u dont want to do it with proper method

1. Easy

Every figure increments another 4 squares-simple.

The cross is increasing in arithamtic sequence ,common difference =4,and first term=1 so,5th term=f+(n-1)d,1+(5-1)4=1+4*4=17

The formula here is 4n - 3 (where n is the number of the Figure). This is because the middle square is always present and the rest is simply 4 x the number of the figure -1 (4 x (n-1)). So the formula = 4( n - 1 ) + 1 (the middle). This simplifies to 4n - 4 + 1 and then 4n - 3. This formula will give us the number of squares in any of these shapes, given the figure number. Here we were asked for Figure 5. (4 x 5) - 3 = 20 - 3 = 17

1, 1+4, 1+4+4, 1+4+4+4, 1+4+4+4+4

always add 4 squares = a= 1 square , b=5 squares c=9 squares d=13 squares (e=17 squares)

For the $n$ th figure, there are always $4n-3$ squares in it.

For the 5th figure, there are $4(5)-3=20-3=17$ squares in it.

just add 4 to the next fig

The $n$ th figure has a quantity of squares equal to $4n - 3$ . Substituting $n = 5$ yields $4(5) - 3 = \boxed {17}$ .

17 Each arm adds another brick, plus the middle brick.

the number of squares on each arm of cross is increasing by 1 in each figure. there were 3 squares in prev fig. so in fig 5 there will be 4 on each arm. total squares= 4Xno.of squares on each arm + 1 central square =>4X4+1=17

If each of the outward drawn parts of cross have $n$ squares

then total squares = $4 \times n+1$

Explanation: $$

1st case: n=0, total = $4 \times 0+1 = 1$

2nd case: n=1, total = $4 \times 1+1 = 5$

3rd case: n=2, total = $4 \times 2+1 = 9$

and so on

Fig.1: 1+ (0 • 4) = 1 █ Fig.2: 1+(1•4) = 5 █ Fig.3: 1+(2•4) = 9 █ Fig.4: 1+(3•4) = 13 █ Fig.5: 1+(4•4) = 17 █ To make it simple, just add 4 squares to the figures.

17. number of squares for Fig. x equals 4n+1, n=x-1.

sum of squares in a given figure is in the form of A.P where a=1 and d=4 :)

4n+1 where n=0,1,2,3.....
4(4)+1=17

U5 = 1 + (5-1)4 = 1+ 16 = 17

Just draw the next figure & you will get it.

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