Let a = min x2+2x+3: x ∈ R and b= lim limitsθ → 0 (1-cos θ/θ2). Then ∑ limitsnr - 0 arbn-r is

Question

Let a = min x2+2x+3: x ∈ R and b= lim limitsθ → 0 (1-cos θ/θ2). Then ∑ limitsnr - 0 arbn-r is (A) (2n+1-1/3.2n) (B) (2n+1+1/3.2n) (C) (4n+1-1/3.2n) (D) (1/2)(2n-1)

Tardigrade Admin · Accepted Answer

a=min (x+1)2+2 = 2 lim limitsθ → 0(2 sin2 (θ/2)/4((θ)2)2) = (1/2), ar bn-r= (2r/2n-r) = 22r-n = (4r/2n), ∑ limitsnr - 0 ar.bn-r = (1/2n) ∑ limitsnr - 0 4r = (1/2n) ((1-4n+1/1-4)) = (4n+1-1/3×2n)

Q. Let a = min ${x^{2} + 2 x + 3 : x \in R}$ and $b = θ \to 0 lim \frac{1 - cos θ}{θ ^{2}}$ . Then $r - 0 \sum n a^{r} b^{n - r}$ is

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

$\frac{2 ^{n + 1} + 1}{3. 2 ^{n}}$

$\frac{4 ^{n + 1} - 1}{3. 2 ^{n}}$

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

Q. Let a = min{x2+2x+3:x∈R} and b=θ→0lim​θ21−cosθ​. Then r−0∑n​arbn−r is

3.2n2n+1−1​

3.2n2n+1+1​

3.2n4n+1−1​

21​(2n−1)

a=min(x+1)2+2=2 θ→0lim​4(2θ​)22sin22θ​​=21​,arbn−r=2n−r2r​=22r−n=2n4r​,r−0∑n​ar.bn−r=2n1​r−0∑n​4r=2n1​(1−41−4n+1​)=3×2n4n+1−1​

Q. Let a = min ${x^{2} + 2 x + 3 : x \in R}$ and $b = θ \to 0 lim \frac{1 - cos θ}{θ ^{2}}$ . Then $r - 0 \sum n a^{r} b^{n - r}$ is

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

$\frac{2 ^{n + 1} + 1}{3. 2 ^{n}}$

$\frac{4 ^{n + 1} - 1}{3. 2 ^{n}}$

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