# 8085 program to find the sum of first n natural numbers

Problem – Write an assembly language program for calculating the sum of first n natural numbers using 8085 microprocessor.

Example –

```Input : 04H
Output : 0AH
as 01+02+03+04 = 10 in decimal => 0AH
```

The formula for calculating the sum of first n natural numbers is .

Algorithm –

1. With n as the input, increment it to obtain n+1.
2. Multiply n with n+1.
3. Divide the product obtained by 2.

In 8085 microprocessor, no direct instruction exists to multiply two numbers, so multiplication is done by repeated addition as 4×5 is equivalent to 4+4+4+4+4 (i.e., 5 times).
Input: 04H
Product: 14H(2010)

Similarly, in 8085 microprocessor, no direct instruction exists to divide two numbers, so division is done by repeated subtraction.
Input: 14H
Keep subtracting 2 from the input till it reduces to 0.
Since subtraction has to be performed 1010 times before 14H becomes 0, the quotient is 1010 => 0AH.

Steps –

1. Load the data from the memory location (201BH, arbitrary choice) into the accumulator
2. Move this data into B
3. Increment the value in the accumulator by one and move it to the register C
4. Initialise the accumulator with 0
5. Multiplcation: Keep adding B to accumulator. The number of times B has to be added is equal to the value of C
6. Initialise B with 00H. B will store the quotient of the division
7. Initialise C with 02H. This is the divisor for the division
8. Division: Keep subtracting C from A till A becomes 0. For each subtraction, increment B by one
9. The final answer is in B. Move it to A. Then store the value of A in 201CH (arbitrary choice again)

2000H LDA 201BH
2001H
2002H
2003H MOV B, A
2004H INR A
2005H MOV C, A
2006H MVI A, 00H
2007H
2009H DCR C
200AH JNZ LOOP1
200BH
200CH
200DH MVI C, 02H
200EH
200FH MVI B, 00H
2010H
2011H LOOP2 INR B
2012H SUB C
2013H JNZ LOOP2
2014H
2015H
2016H MOV A, B
2017H STA 201CH
2018H
2019H
201AH HLT

Store the value of n in 201BH. The sum can be found at 201CH.