SOLVING PROBLEMS WITH PATTERNS

Example 1 :

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

(a) 8^th figure

(b) n^th figure

Solution :

By observing the figures

Number of matchstick in the 1^st figure = 4.

Number of matchstick in the 2^nd figure = 10.

Number of matchstick in the 3^rd 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 n^th 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 n^th 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) S_n

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,

S_n = n/(n + 1)

(where n is natural number)

(a)  

(b) (i) If n = 10,

S₁₀ = 10/(10 + 1)

S₁₀ = 10/11

S_n = n/(n + 1)

(ii) S_n = n/(n + 1)

Example 3 :

Consider the pattern :

S₁ = 1²

 S₂ = 1² + 2²

 S₂ = 1² + 2² + 3², …………

a) Check that the formula

S_n = [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,

S_n = [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.

S_n = [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

N_n = [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,

N_n = [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.

S_n = 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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