Question

The sum of n terms of (1-1/n)+(2-4/n)+(3-9/n) +n...​

Answer 1

Answer:

We have

Sm=1+2(1+1/n)+3(1+1/n)2+4(1+1/n)3+...…..+m(1+1/n)m−1 …………(A)

multiplying each term by (1+1/n) we get,

Sm(1+1/n)=1(1+1/n)+2(1+1/n)2+3(1+1/n)3+4(1+1/n)4+...……+(m−1)(1+1/n)m−1+m(1+1/n)m ……………(B)

Subtracting (B) from (A), we get

−nSm=1+(1+n1)+(1+1/n)2+(1+1/n)3+...…..+(1+1/n)m−1−m(1+1/n)m

=1+n1−1(1+1/n)m−1−m(1+1/n)m

(or)

Sm=mn(1+1/n)m−n2((1+1/n)m−1)

and if m=n

Sn=n2(1+1/n)n−n2((1+1/n)n−1)

Sn=n2

∴ So, the answer is A. n2.