"indukshun" meen yu sho its true for n=1 or 2
then if true for n, it must be tru for (n+1)
sum av sequens...1+2+3+4+5+...=n*(n+1)/2
for n=1, sum=1 & formula is 1*2/2=1, so its tru for n=1
then chaenj n tu n+1...formula giv (n+1)*(n+2)/2...(n^2+3n+2)/2
this shood =[n*(n+1)/2] +(n+1)...yu add 1 number tu prior sum
=[n^2 +n +2n+2]/2
=(n^2 +3n+1)/2
we got the formula us spekt