need to solve recurrence t(n)=t(n-1)+n, for n>1, t(1)=a; where a is some constant of type double Enter a and n

1. Find t(a, n) using recursion What is complexity of your solution, when you use recursion?

2. find t(a,n)==f(a,n) in O(1) complexity Provide code for both solutions. for n<1, t(a,n) is not defined, so throw exception if user enters not number, and throw exception if user enters number less than 1 catch both exceptions and ask user to re-enter n value

Intended output:

You need to solve recurrence t(n)=t(n-1)+n, for n>1, t(1)=a;

where a is some constant of type double

Enter a and n 1. Find t(a, n) using recursion

What is complexity of your solution, when you use recursion?

2. find t(a,n)==f(a,n) in O(1) complexity

Provide code for both solutions.

for n<1 t(a,n) is not defined, so throw exception if user enters

not number, and throw exception if user enters numberless than 1

catch both exceptions and ask user to re-enter n value

t(1)=a Enter constant double a = w

you entered not double type value

t(1)=a Enter constant double a = r

t(1)=a Enter constant double a = 0.5

Enter int n =0

you entered 0 number is less than 1

t(1)=a Enter constant double a = 4

Enter int n =-3

you entered -3 number is less than 1

t(1)=a Enter constant double a = 0.5

Enter int n =8

recursive t(0.5,8)=35.5

complexity O(1) f(0.5,8)=35.5

Do you want to repeate? y/n: y

t(1)=a Enter constant double a = r

t(1)=a Enter constant double a = 1

Enter int n =5

recursive t(1,5)=15

complexity O(1) f(1,5)=15

Do you want to repeat? y/n: n