Evaluate the following limits: _ x 1 (x+3 x^2+5 x^3+ +(2 n-1) x^n-n^2 x-1 )

_ x 1 (x+3 x^2+5 x^3+ +(2 n-1) x^n-n^2 x-1 ) Consider 1+3+5+ +(2 n-1) = _ r=1 ^n(2 r-1) =2 _ r=1 ^n r- _ r=1 ^n 1 =2 n(n+1) 2 -n =n(n+1)-n =n^2+n-n =n^2 n^2=1+3+5+ +(2 n-1) . Required limit = _ x 1 [x+3 x^2+5 x^3+ +(2 n-1) x^n ]-[1+3+5+ +(2 n-1)] x-1 = _ x 1 (x-1)+ (3 x^2-3 )+ (5 x^2-5 )+ +(2 n-1) x^n-(2 n-1) x-1 = _ x 1 [x-1 x-1 +3 (x^2-1 ) x-1 +5 (x^3-1 ) x-1 + +(2 n-1) (x^n-1 ) x-1 ] = _ x 1 (x^1-1^1 x-1 )+3 _ x 1 (x^2-1^2 x-1 )+5 _ x 1 (x^3-1^3 x-1 ) + +(2 n-1) _ x 1 (x^n-1^n x-1 ) =1(1)^0+3(2)(1)^1+5(3)(1)^2+ +(2 n-1) n(1)^ n-1 =1(1)+3(2)+5(3)+ +(2 n-1) n = _ r=1 ^n(2 r-1) r = _ r=1 ^ (2 r^2-r ) =2 _ r=1 ^n r^2- _ r=1 ^n r =2 n(n+1)(2 n+1) 6 - n(n+1) 2 =n(n+1) (2 n+1 3 -1 2 ) =n(n+1) (4 n+2-3 6 )=n(n+1)(4 n-1) 6

Evaluate the following limits: _ x 1 (x+3 x^2+5 x^3+ +(2 n-1) x^n…

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Question

Evaluate the following limits:
$\lim _{x \rightarrow 1}\left(\frac{x+3 x^2+5 x^3+\cdots \cdots \cdots \cdots \cdots+(2 n-1) x^n-n^2}{x-1}\right)$

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Answer

$\lim _{x \rightarrow 1}\left(\frac{x+3 x^2+5 x^3+\cdots+(2 n-1) x^n-n^2}{x-1}\right)$
Consider
$1+3+5+\ldots+(2 n-1)$
$=\sum_{r=1}^n(2 r-1)$
$=2 \sum_{r=1}^n r-\sum_{r=1}^n 1$
$=2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}-\mathrm{n}$
$=n(n+1)-n$
$=n^2+n-n$
$=\mathrm{n}^2$
$\therefore \quad n^2=1+3+5+\ldots+(2 n-1) \text {. }$
$\therefore \quad \text { Required limit }$
$=\lim _{x \rightarrow 1} \frac{\left[x+3 x^2+5 x^3+\cdots+(2 n-1) x^n\right]-[1+3+5+\cdots+(2 n-1)]}{x-1}$
$=\lim _{x \rightarrow 1} \frac{(x-1)+\left(3 x^2-3\right)+\left(5 x^2-5\right)+\cdots+(2 \mathrm{n}-1) x^n-(2 \mathrm{n}-1)}{x-1}$
$=\lim _{x \rightarrow 1}\left[\frac{x-1}{x-1}+\frac{3\left(x^2-1\right)}{x-1}+\frac{5\left(x^3-1\right)}{x-1}+\cdots+\frac{(2 n-1)\left(x^n-1\right)}{x-1}\right]$
$=\lim _{x \rightarrow 1}\left(\frac{x^1-1^1}{x-1}\right)+3 \lim _{x \rightarrow 1}\left(\frac{x^2-1^2}{x-1}\right)+5 \lim _{x \rightarrow 1}\left(\frac{x^3-1^3}{x-1}\right)$
$+\cdots+(2 n-1) \lim _{x \rightarrow 1}\left(\frac{x^n-1^n}{x-1}\right)$
$=1(1)^0+3(2)(1)^1+5(3)(1)^2+\ldots+(2 n-1) n(1)^{n-1}$
$=1(1)+3(2)+5(3)+\ldots+(2 n-1) n$
$=\sum_{r=1}^n(2 \mathrm{r}-1) \mathrm{r}$
$=\sum_{r=1}^{\infty}\left(2 r^2-r\right)$
$=2 \sum_{r=1}^n r^2-\sum_{r=1}^n r$
$=2 \cdot \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
$=n(n+1)\left(\frac{2 n+1}{3}-\frac{1}{2}\right)$
$=n(n+1)\left(\frac{4 n+2-3}{6}\right)=\frac{n(n+1)(4 n-1)}{6}$

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