Sum calculator in MIPS - recursion

Question

I am fairly new to MIPS and am trying to write a code which calculates the sum of an integer which was entered by the user. I am using recursion in my code but am not sure if theres an easier way to do this. My code runs fairly well for the most part however it keeps adding 1 to the final sum. For example, sum of 10 is 55 but it gives me 56.
.data
#messages
sumMessage: .asciiz "\Enter a number to find its Sum: "
sumMessage1: .asciiz "\nSum is: "
sumInput: .word 0
sumAns: .word 0
.text
.globl main
main:
#read promtMessage
li $v0, 4
la $a0, sumMessage
syscall
#take input
li $v0,5
syscall
sw $v0, sumInput#storing input
#call sumCalc function
lw $a0, sumInput
jal sumCalc
sw $v0, sumAns#returns value from function
#display sum
li $v0, 4
la $a0, sumMessage1
syscall
li $v0, 1
lw $a0, sumAns
syscall
#end main
li $v0, 10
syscall
#sumCalc Function
.globl sumCalc
sumCalc:
subu $sp, $sp, 8
sw $ra, ($sp)#storing value of returning address in the stack
sw $s0, 4($sp)#4 bites apart from the value stored above in the stack
#base case
li $v0, 1
beq $a0, 0, exit
#find factorial n-1
move $s0, $a0
sub $a0, $a0, 1
jal sumCalc
add $v0, $s0, $v0
exit:
lw $ra, ($sp)
lw $s0,, 4($sp)
addu $sp, $sp, 8
jr $ra
Any help would be appreciated!

Answer 1

Using recursion here makes the processing cumbersome and of little interest because we can calculate the sum directly. The sum of an integer is given to us using the following mathematical formula: n * (n + 1) / 2 or else n * (n + 1) then a right shift which is comparable to a division by 2.
.data
#messages
sumMessage: .asciiz "\Enter a number to find its Sum: "
sumMessage1: .asciiz "\nSum is: "
sumInput: .word 0
sumAns: .word 0
.text
.globl main
main:
#read promtMessage
li $v0, 4
la $a0, sumMessage
syscall
#take input
li $v0,5
syscall
sw $v0, sumInput#storing input
#call sumCalc function
lw $a0, sumInput
jal sumCalc
sw $s2, sumAns#returns value from function
#display sum
li $v0, 4
la $a0, sumMessage1
syscall
li $v0, 1
lw $a0, sumAns
syscall
#end main
li $v0, 10
syscall
#sumCalc Function
.globl sumCalc
sumCalc:
subu $sp, $sp, 8
sw $ra, ($sp)#storing value of returning address in the stack
sw $s0, 4($sp)#4 bites apart from the value stored above in the stack
li $s2,1 # storing the final result
move $s3,$v0
addi $s3,$s3,1 # $s3 = $v0+1
mul $s2,$v0,$s3 # $s2 = $v0 * ($v0 +1)
div $s2,$s2,2 # $s2 = $s2/2
exit:
lw $ra, ($sp)
lw $s0,4($sp)
addu $sp, $sp, 8
jr $ra