displaystyle limx → 1(x+x2+...+xn-n/x-1) is

Question

displaystyle limx → 1(x+x2+...+xn-n/x-1) is (A) n (B) (n+1/2) (C) (n(n+1)/2) (D) (n(n-1)/2)

Tardigrade Admin · Accepted Answer

We have displaystyle limx → 1 (x+x2+...+xn-n/x-1) = displaystyle limx → 1 (x+x2+...+xn-1-1-1-...-1/x-1) = displaystyle limx → 1 ((x-1)+(x2-1)+(x3-1)+...+(xn-1)/x-1) = displaystyle limx → 1 1 + (x+ 1) + (x2 + x +1) + ... =1+2+3+...+n =(n(n+1)/2)

Q. $x \to 1 lim$ $\frac{x + x ^{2} + ... + x ^{n} - n}{x - 1}$ is

$n$

$\frac{n + 1}{2}$

$\frac{n ( n + 1 )}{2}$

$\frac{n ( n - 1 )}{2}$

Q. x→1lim​x−1x+x2+...+xn−n​ is

n

2n+1​

2n(n+1)​

2n(n−1)​

We have x→1lim​{x−1x+x2+...+xn−n​} =x→1lim​{x−1x+x2+...+xn−1−1−1−...−1​} =x→1lim​{x−1(x−1)+(x2−1)+(x3−1)+...+(xn−1)​} =x→1lim​{1+(x+1)+(x2+x+1)+...} =1+2+3+...+n =2n(n+1)​

Q. $x \to 1 lim$ $\frac{x + x ^{2} + ... + x ^{n} - n}{x - 1}$ is

$n$

$\frac{n + 1}{2}$

$\frac{n ( n + 1 )}{2}$

$\frac{n ( n - 1 )}{2}$