If a= min x2+4 x+5, x ∈ R and b= displaystyle lim θ arrow 0 (1- cos 2 θ/θ2), then the value of displaystyle∑r=0n ar ⋅ bn-r is

Question

If a= min x2+4 x+5, x ∈ R and b= displaystyle lim θ arrow 0 (1- cos 2 θ/θ2), then the value of displaystyle∑r=0n ar ⋅ bn-r is (A) (2n+1-1/4 ⋅ 2n) (B) 2n+1-1 (C) (2n+1-1/3 ⋅ 2n) (D) None of these

Tardigrade Admin · Accepted Answer

x2+4 x+5=(x+2)2+1 ≥ 1 . text So, a=1 Also, b= displaystyle lim θ arrow 0 (1- cos 2 θ/θ2) = displaystyle lim θ arrow 0 (2 sin 2 θ/θ2)=2 ∴ displaystyle∑r=0n ar ⋅ bn-r=bn+a bn-1+a2 bn-2+ ldots+an =(bn[1-((a/b))n+1]/1-(a)b) =(2n[1-((1/2))n+1]/1-(1)2) =(2n+1(2n+1-1)/2n+1) =(2n+1-1)

Q. If $a = min {x^{2} + 4 x + 5, x \in R}$ and $b = θ \to 0 lim \frac{1 - cos 2 θ}{θ ^{2}}$ , then the value of $r = 0 \sum n a^{r} \cdot b^{n - r}$ is

$\frac{2 ^{n + 1} - 1}{4 \cdot 2 ^{n}}$

$2^{n + 1} - 1$

$\frac{2 ^{n + 1} - 1}{3 \cdot 2 ^{n}}$

None of these

Q. If a=min{x2+4x+5,x∈R} and b=θ→0lim​θ21−cos2θ​, then the value of r=0∑n​ar⋅bn−r is

4⋅2n2n+1−1​

2n+1−1

3⋅2n2n+1−1​

None of these

x2+4x+5=(x+2)2+1≥1. So, a=1 Also, b=θ→0lim​θ21−cos2θ​ =θ→0lim​θ22sin2θ​=2 ∴r=0∑n​ar⋅bn−r=bn+abn−1+a2bn−2+…+an =1−ba​bn[1−(ba​)n+1]​ =1−21​2n[1−(21​)n+1]​ =2n+12n+1(2n+1−1)​ =(2n+1−1)

Q. If $a = min {x^{2} + 4 x + 5, x \in R}$ and $b = θ \to 0 lim \frac{1 - cos 2 θ}{θ ^{2}}$ , then the value of $r = 0 \sum n a^{r} \cdot b^{n - r}$ is

$\frac{2 ^{n + 1} - 1}{4 \cdot 2 ^{n}}$

$2^{n + 1} - 1$

$\frac{2 ^{n + 1} - 1}{3 \cdot 2 ^{n}}$