0
an=an-1
To use this formula, you start of with a value on the first term, but theoretical, it'd turn out like this:
a1=a1-1=a0
a2=a2-1=a1
a3=a3-1=a2
a4=a4-1=a3
a5=a5-1=a4
Where a0 would be your starting term (this formula is based on the previous term, and that's why you must have a value to start off with).
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Codings for coprime number in c?
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
What is the value of the expression n1?
the value of the exponent n1
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The sum of the first 10 positive integers, using the formula N1 + (N1 + 1) + ... + N2 = N2 * (N2 + 1) / 2 - (N1 - 1) * N1 / 2 is: 55
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You subtract the exponents. N30 - N1 = N30 - 1 = N29.You subtract the exponents. N30 - N1 = N30 - 1 = N29.You subtract the exponents. N30 - N1 = N30 - 1 = N29.You subtract the exponents. N30 - N1 = N30 - 1 = N29.
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Codings for coprime number in c?
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
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P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n
Find LCM and HCF of two given number by using for loop in c?
#include<stdio.h> int main(){ int n1,n2; printf("\nEnter two numbers:"); scanf("%d %d",&n1,&n2); while(n1!=n2){ if(n1>=n2-1) n1=n1-n2; else n2=n2-n1; } printf("\nGCD=%d",n1); return 0; }
What is the value of the expression n1?
the value of the exponent n1
What is the value of the expression of n1?
the value of the exponent n1
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void main() { int i; float n1,n2; abc: printf("Enter two nos "); scanf("%f%f",&n1,&n2); printf("\n %f + %f = %f " ,n1,n2,n1+n2); printf("\n %f - %f = %f " ,n1,n2,n1-n2); printf("\n %f x %f = %f " ,n1,n2,n1*n2); printf("\n %f / %f = %f " ,n1,n2,n1/n2); printf("\npress 5 to make another calculation"); scanf("%d",&i); if (i==5) goto abc; }
What is the sum of the 10 positive integers?
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