Find first five terms of the sequence. a_( n) = ( 1)/( 1 + a_( n - 1 ) ) and a_( 1) = 1

Using recursive formula to find the next five terms

a_1  = 1

Second term 

n = 2

a_2   =  ( 1)/(1 + a_( 2 - 1) )

a_( 2)  =  1/(1 +〖 a〗_( 1) )

a_2  =( 1)/(1 + 1)   =  ( 1)/( 2)

Third term 

n = 3

a_3   =  ( 1)/(1 + a_( 3 - 1) )

a_3   =  1/(1 + 〖a 〗_2 )

a_3  =( 1)/(1 + 1/( 2))   =   2/3

Fourth term 

n = 4

a_4  =1/(1 + a_( 4 - 1) )

a_4  =( 1)/(1 +〖 a〗_3 )

a_4  =1/(1+2/3)   =   3/5

Fifth term 

n=5

a_5  =  ( 1)/(1 + 〖a 〗_(5 - 1) )

a_5   =  ( 1)/(1 +〖 a〗_( 4) )

a_5  =  ( 1 )/(1 + ( 3)/( 5))   =   ( 5)/( 8)

The first five terms are 1,1/2,2/3,3/5,5/8. 

Question No 68

Write the sum using sigma notation.

2 + 5 + 8 + ..…… … …  29

Considering the geometric sequences for the sigma notations, 

a_( n)  = 29

a_( 1)  = 2 

a_( n)=  a_( 1)   + d ( n - 1 )

d = 3 

29=  2 +3 ( n - 1 )

29 =  2 + 3n - 3

30 = 3 n 

10 = n 

Arithmetic series in sigma notation,

∑_(i=1)^10▒〖3n - 1〗  

Question No 70

Write the sum using sigma notation.

( 1)/( 2 ln 2)  -  ( 1)/(3 ln 3) +   ( 1)/(4 ln 4)-  (1 )/(5 ln 5)… … .( 1)/(100 ln 100)

index variables  =  n = 2 

a_1  =  ( 1)/(2 ln⁡〖 2〗 )

sign =  (-1)^( n)

〖a 〗_n=   ( 1)/(100 ln 100)

n  =  100

∑_(n = 2)^100▒( (-1)^n)/(n 〖ln 〗⁡(n) )

Question No 72

Write the sum using sigma notation.

√1/1^( 2)    +   ( √2)/〖2 〗^2    +  ( √3)/3^( 2)   + … … … .+  ( √n)/〖n 〗^2 

a_1  =√1/1^2   

sign =  (1)^n

a_n= √( n)/n^( 2) 

∑_(n = 1)^n▒( √( n))/n^2 

Question No 72

Telephones poles are being stored in a pile with 25 poles in the first, 24 in the second, and so on. If there are 12 layers, how many telephone poles does the pile contain?

number of poles in first layer= a_1=25 

total number of poles =12

The number of poles decreased by 1 as we move forward 

number of poles decreased= d= 1 

sum =  12/2 (a_1+a_n )

a_12  =  25+2 (-1)

a_12  =  25+11 (-1)

a_12  =  14 

Arithmetic sequences in the first term a and common differences as -d. the number of poles in the nth layer.

sum =  12/2 (25+14)

number of poles= 234 

There will be 234 telephone poles in the selected area.

Question No: 76

When an object is allowed to fall freely near the surface of the earth, the gravitational pull is such that the object falls 16 ft in the first second, 48 ft in the next second, 80 ft in the next second, and so on

Total distance a ball falls in 6 s.

With each second, the distance of ball is increasing and it will use arithmetic regression with increasing trends. The distance covered by freely falling object in first second is 16 ft, in next second it is 48 ft, in the next second it will be 80 ft. The total flight time of the ball is 6 s. the nth partial sum of the arithmetic sequence becomes as follows, 

S_n  =n/2  [2a+  (n-1)d]  

The gravitational pull on the ball at different distance and in every second is changing. The first, second, and third terms of arithmetic progression will be 16 ft, 48 ft, and 80 ft respectively. 

a=  16 

d =  32 

a_n  =  16 +32 (n-1 )

a_n  =  16 +32 (5 )

a_6=  16+  160  =  176

Find a formula for the total distance a ball falls in n second.

formula for the total distance covered by ball 

S_n  =  n/2  [ 2×16+32 (n-1)]

S_n=n/2×32 n

S_n  =  16 n^2  

formula for the total distance=16 n^2