the sum of 1+2+3+4+5+6 and so on
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this is done by using the formula :- a + (n-1)d
where a is first number and d is the difference between first two numbers and 'n' is no. of digits that is upto how many terms
singhabhay03:
ooooooooops
it is = n/2(2a+(n-1)d)
by using arithmetic progression
yesssssssssssss
sum of all nth natural number
hmmm
you are right
thnx
dude
Answered by
2
hlo mate.
here is your answer.
__________________
The sum of 1+2+3+4+5+6. ...and so on. ..___>>..this can be done by using the formula a+(n-1)d.
_______________
hope it helps you
here is your answer.
__________________
The sum of 1+2+3+4+5+6. ...and so on. ..___>>..this can be done by using the formula a+(n-1)d.
_______________
hope it helps you
tnx for brainliest
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