The general term of $(2 a - 3 b)^{- 1 / 2}$ is

\frac{1.3.5 \dots \dots (2 r - 5) 1}{r! \sqrt{2 a}} (\frac{3 b}{2 a})^{r + 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1..5 \dots . . (2 r - 5) 1}{r! \sqrt{2} a} (\frac{3 b}{4 a})^{r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1.3.5 \dots . . . \cdot (2 r - 1) 1}{2^{r} r! \sqrt{2 a}} (\frac{3 b}{2 a})^{r}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1.3.5 \dots \dots (2 r + 5)}{r!} (\frac{1}{\sqrt{2 a}}) (\frac{3 b}{4 a})^{r - 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{1.3.5 \dots . . . \cdot (2 r - 1) 1}{2^{r} r! \sqrt{2 a}} (\frac{3 b}{2 a})^{r}$
$(2 a - 3 b)^{- 1 / 2}$ $= \frac{1}{\sqrt{2 a}} {(1 - \frac{3 b}{2 a})}^{- 1 / 2}$
Hence general term would be
$\frac{1}{\sqrt{2 a}} (\frac{\frac{1}{2} \frac{1}{2} (\frac{1}{2} + 1) . . . (\frac{1}{2} + (r - 1))}{r!} (\frac{3 b}{2 a})^{r}$
$= \frac{1}{\sqrt{2 a}} (\frac{1.3.5... (2 r - 1)}{2^{r} r!}) (\frac{3 b}{2 a})^{r}$

Suggest Corrections

Similar questions

If $f (x) = l i m n \to \infty \sum_{r - = 1}^{n} \frac{r x}{1.3.5 \dots (2 r + 1)}$ , then $\int_{0}^{3} [f (x)] d (x - [x])$ is equal to (where [.] denotes greatest integer function)

Q. Solve :

Σ_{r = 1}^{n} (2 r - 1) (2 r + 1)

Q. Evaluate:

\sum_{r = 1}^{\infty} {tan}^{- 1} (\frac{2}{1 + (2 r + 1) (2 r - 1)})

Q. Show that the highest power of

2

contained in

2^{r} - 1 - ----- -

2^{r} - r - 1

50 \sum r = 1 [\frac{1}{49 + r} - \frac{1}{2 r (2 r - 1)}] =

Join BYJU'S Learning Program

The general term of (2a−3b)−1/2 is

The correct option is C 1.3.5…...⋅(2r−1)12rr!√2a(3b2a)r(2a−3b)−1/2 =1√2a(1−3b2a)−1/2Hence general term would be1√2a(1212(12+1)...(12+(r−1))r!(3b2a)r=1√2a(1.3.5...(2r−1)2rr!)(3b2a)r

The general term of $(2 a - 3 b)^{- 1 / 2}$ is