Really tough question #9-10

Question

Consider a chessboard, which has a total of 64 squares. One grain of wheat is placed on the 1st square, 2 grains on the 2nd square, 4 grains on the 3rd square, 8 grains on the 4th square, and so on.

How many grains of wheat will be placed on square number 19?

Assuming that each grain of wheat weighs 1/7000 pound, what will be the total weight (in pounds) of the grains on the entire chessboard when 19 squares are filled?

Answer 1

this is a geometric sequence

A(n)=a*(r)^(n-1) where r=common ratio, n=# of terms, a= first term

A(19)=1(2)^(19-1)

A(19)=2^(18)

A(19)=262,144 grains on square #19

find the sum of the grains of wheat

S=a(r^n-1)/(r-1)

S=1(2^19-1)/(2-1)

S=(524,288-1)

S=524,287 grains of wheat

weight of the grains is (524,287)(1/7000)=524,287/7000

524,287/7000=74.89814 pounds

Answer 2

Jesse,

This is just a geometric sequence, where the previous term is multiplied by 2.  Well isn't multiplying by 2, the same as using powers of 2?  Yep!  2⁰ = 1 , 2¹ = 2, 2² = 4........

a₁ = 1

a₂ = 2

a₃ = 4

So,

a_n = 2(a_n-1)   or a_n = 2^n-1

So on the 19th square:

a₁₉ = 2^19-1

a₁₉ = 2¹⁸

a₁₉ =  262144   

 

262144 grains of rice are on the 19th square.

 

For the 2nd part, this is a sum of a sequence that has a starting term of 1/7000 lb, and a common ratio of 2, which can be expressed using the formula:

 

n(∑)i=1 a_i = a((1-rⁿ)/(1-r))  ; Where n is the number of terms, i is the index, a is the first term, and r is the common ratio.

 

19(∑)i=1 a₁₉ = (1/7000)((1-2¹⁹)/(1-2)) 

 

19(∑)i=1  a₁₉ = (1/7000)((1-524288)/-1)

19(∑)i=1  a_{19 = (1/7000)(-524287/-1)}

19(∑)i=1  a₁₉ = (1/7000)(524287)

19(∑)i=1  a₁₉ = 524287/7000

19(∑)i=1  a₁₉ ≈ 74.9 lbs