Question

Pad 8X(1+2+3+ ... +499)+1=? answer


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Answer 1

Answer:

998001

Step-by-step explanation:

8×(1+2+3+4+....+499)+1

here we can see that (1+2+3+4+....+499) are in a.p (arithmetic progression) as d(common difference)=1

so by solving this by a.p we get,

Sn=n/2[a+l]=499/2[1+499]=124750

n=number of terms=499

a=first term of the a.p=1

l=last term of the a.p=499

hence,

8×(124750)+1=998001