## SOLVING PROBLEMS WITH PATTERNS

Example 1 :

For the following matchstick pattern, find the number of matches M required to make the

(a)  8th figure

(b)  nth figure

Solution :

By observing the figures

Number of matchstick in the 1st figure  =  4

Number of matchstick in the 2nd figure  =  10

Number of matchstick in the 3rd figure  =  16

Number of matchsticks in every figure is 2 less than the multiple of 6.

So, creating formula for this

number of matchstick in nth figure  =  6n-2

(a)  using the formula

n = 8

Number of match sticks in 8th figure  =  6(8)-2

=  48-2

=  46

(b)  Number of matchstick in nth figure  =  6n-2

Example 2 :

Consider the pattern :

S1  =  1/1×2

S2  =  1/1×2 + 1/2×3

S3  =  1/1×2 + 1/2×3 + 1/3×4

………….......

.....................

a)  Find the values of S1, S2, S3, and S4

b) write down the value of :

(i) S10   (ii) Sn

Solution :

S1  =  1/(1 x 2)  =  1/2

S2  =  1/(1 x 2) + 1/(2 x 3)   =  1/2 + 1/6  ==>  4/6  =>  2/3

S3  =  1/(1x2) + 1/(2x3) + 1/(3x4)

=  1/2 + 1/6 + 1/12

=  (6+2+1)/12

=  9/12

=  3/4

Observing the results it is in the form n/(n+1).

So,

Sn  =  n/n+1

(where n is natural number)

(a)

 If n  =  1S1  =  1/1+1S1  =  1/2If n  =  2S2  =  2/2+1S2  =  2/3 If n  =  3S3  =  3/3+1S3  =  3/4If n  =  4S4  =  4/4+1S4  =  4/5

(b) (i)  If n  =  10

S10  =  10/10+1

S10  =  10/11

Sn  =  n/n+1

(ii)  Sn  =  n/n+1

Example 3 :

Consider the pattern :

S1  =  12

S2  =  12 + 22

S2  =  12 + 22 + 32, …………

a)  Check that the formula

Sn  =  n(n+1) (2n+1)/6

is correct for n  =  1, 2, 3 and 4

b) Assuming the formula in a is always true, find the sum of

12+22+32+42+52+ ………… + 1002

which is the sum of the squares of the first one hundred integers.

Solution :

(a)  When n  =  1,

Sn  =  n(n+1) (2n+1)/6

S1  =  1(1+1) (2+1)/6

S1  =  1

It can be written as 12

when n  =  2,

S2  =  2(2+1) (4+1)/6

S2  =  5

It can be written as 12 + 22

when n  =  3,

S3  =  3(3+1) (6+1)/6

S2  =  84

It can be written as 12 + 22 + 32

Hence its verified.

(ii)  Given :

12+22+32+42+52+ ………… + 1002

Here, n  =  100

Sn  =  n(n+1) (2n+1)/6

S100  =  100(100+1) (200+1)/6

S100  =  338350

Example 4 :

Consider the pattern :

N1  =  13

N2  =  13 + 23

N3  =  13 + 23 + 33, …………

a)  Verify that the formula

Nn  =  n2(n+1)2/4

is correct for n  =  1, 2, 3 and 4

b) Use the above formula to find the sum of

13 + 23 + 33 + 43 + ………… + 503

c) Find the sum : 23 + 43 + 63 + 83 + ………… + 1003

Solution :

(a)  When n  =  1,

Nn  =  n2(n+1)2/4

N1  =  12(1+1)2/4

N1  =  1

It can be written as 13

when n  =  2,

N2  =  22(2+1)2/4

N2  =  9

It can be written as 13 + 23

when n  =  3,

N3  =  32(3+1)2/4

N3  =  36

It can be written as 13 + 23 + 33

Hence, its verified.

(b)   13 + 23 + 33 + 43 + ………… + 503

here, n  =  50

Sn  =  n2(n+1)2/4

S50  =  (50)2(50+1)2/4

S50  =  1625625

c) 23 + 43 + 63 + 83 + ………… + 1003

Factoring 23 from the series.

=  23 (13 + 23 + 33 + 43 + ………… + 503)

=  23(1625625)

=  8(1625625)

=  13005000

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