Question: If $\sum_{r = 1}^{k}\cos^{- 1}$β<sub>r</sub> =$\frac{k\pi}{2}$for any k ≥ 1 and A =\(\sum_{r = 1...

Question

If $\sum_{r = 1}^{k}\cos^{- 1}$ β_r = $\frac{k\pi}{2}$ for any k ≥ 1 and A = $\sum_{r = 1}^{k}{(\beta_{r})^{r}}$ , Then $\frac{(1 + x^{2})^{1/3} - (1 - 2x)^{1/4}}{x + x^{2}}$ =

Answer 1

Solution

For k = 1, cos^–1 β₁ = $\frac { \pi } { 2 }$ ⇒ β₁ = cos $\frac { \pi } { 2 }$ = 0

For k = 2, cos^–1 β₁ + cos^–1β₂ = π ⇒ cos^–1 β₂ = $\frac { \pi } { 2 }$ ⇒ β₂ = 0

So by induction it can be shown that β_j = 0, j = 1……..k.

Hence A = $\sum _ { r = 1 } ^ { k }$ (β_r)^r = 0.

Thus required limit

= $\lim _ { x \rightarrow 0 }$

= $\lim _ { x \rightarrow 0 }$ = $\frac { 1 } { 2 }$

Question: If \(\sum_{r = 1}^{k}\cos^{- 1}\)β<sub>r</sub> =\(\frac{k\pi}{2}\)for any k ≥ 1 and A =\(\sum_{r = 1...

Solution