The value of undersetx arrow 1lim displaystyle ∑ r = 110 (xr - 1r/2 (x - 1)) is equal to

Question

The value of undersetx arrow 1lim displaystyle ∑ r = 110 (xr - 1r/2 (x - 1)) is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

We have, undersetx arrow 1lim displaystyle ∑ r = 110 (xr - 1r/2 (x - 1)) =(1/2) displaystyle ∑ r = 110r⋅ 1r - 1=(1 + 2 + 3 + ldots + 10/2)=(10 (10 + 1)/4)=(55/2)

Q. The value of $x \to 1 l im r = 1 \sum 10 \frac{x ^{r} - 1 ^{r}}{2 ( x - 1 )}$ is equal to

We have, $x \to 1 l im r = 1 \sum 10 \frac{x ^{r} - 1 ^{r}}{2 ( x - 1 )}$
$= \frac{1}{2} r = 1 \sum 10 r \cdot 1^{r - 1} = \frac{1 + 2 + 3 + \dots + 10}{2} = \frac{10 ( 10 + 1 )}{4} = \frac{55}{2}$

Q. The value of x→1lim​r=1∑10​2(x−1)xr−1r​ is equal to

We have, x→1lim​r=1∑10​2(x−1)xr−1r​ =21​r=1∑10​r⋅1r−1=21+2+3+…+10​=410(10+1)​=255​

Q. The value of $x \to 1 l im r = 1 \sum 10 \frac{x ^{r} - 1 ^{r}}{2 ( x - 1 )}$ is equal to

We have, $x \to 1 l im r = 1 \sum 10 \frac{x ^{r} - 1 ^{r}}{2 ( x - 1 )}$
$= \frac{1}{2} r = 1 \sum 10 r \cdot 1^{r - 1} = \frac{1 + 2 + 3 + \dots + 10}{2} = \frac{10 ( 10 + 1 )}{4} = \frac{55}{2}$