Question

(1-1/n)+(1-2/n)+(1-3n)+... upto n terms=?

Answer 1

So it can be simply written as,
=(1-1/n)+(1-2/n)+(1-3/n)+..........upto n terms.
=(1+1+1+1+1..... n terms)-(1/n+2/n+3/n+........n terms)
=n-[(1+2+3+4+5+......+n)/n]
=n-[n(n+1)/2n]
=n-[(n+1)/2]
=(2n-n-1)/2
thus, answer is (n-1)/2

Thanks.
Tripathy.