11-2 -1-2-3 - 1-3-4 - 1-24-25 -

The correct option is D 4 → General term, T_r = 1/√(r+1)+√(r) ⇒ T_r = √(r+1) √(r)/(r+1) r= √(r+1) √(r) ⇒ S = ∑ T_r = √(2) √(1) + ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Solving Using Quadratic Formula When D>0

11+√2 +1√2+√3...

Question

$\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + . . . . . . . . + \frac{1}{\sqrt{24} + \sqrt{25}}$ =..............

1

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

2

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

3

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

4

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

Open in App

Solution

The correct option is D $4$
$\to$ General term, $T_{r} = \frac{1}{\sqrt{r + 1} + \sqrt{r}}$
$\Rightarrow T_{r} = \frac{\sqrt{r + 1} - \sqrt{r}}{(r + 1) - r} = \sqrt{r + 1} - \sqrt{r}$
$\Rightarrow S = \sum T_{r} = \sqrt{2} - \sqrt{1}$
$+ \sqrt{3} - \sqrt{2} = \sqrt{25} - \sqrt{1}$
$\sqrt{4} - \sqrt{3} = 5 - 1$
$= 4$
$\sqrt{25} - \sqrt{24} \Rightarrow (D)$

Suggest Corrections

Similar questions

Q. If

{(1 + x + x^{2})}^{25} = \sum_{r = 0}^{50} a_{r} x^{r},

then

\sum_{r = 0}^{12} a_{4 r}

equals

The sum of the first 25 terms of the series

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

Q. Observe the following patterns

1 = \frac{1}{2} (1 \times (1 + 1))

1 + 2 = \frac{1}{2} (2 \times (2 + 1))

1 + 2 + 3 = \frac{1}{2} (3 \times (3 + 1))

1 + 2 + 3 + 4 = \frac{1}{2} (4 \times (4 + 1))

and find the value of each of the following

1 + 2 + 3 + 4 + 5 + . . . . + 50

Q. Express the following rations in the simplest form.

1) 6 : 8

2) 21 : 24

3) 33 : 77

4) 25 : 125

5) \frac{2}{3} : \frac{3}{5}

6) \frac{4}{5} : \frac{3}{8}

7) 1 \frac{1}{2} : 4 \frac{1}{2}

8) \frac{1}{3} : 2

Simplify : 25-1/2{5+4-(3+2-1+3)}

Join BYJU'S Learning Program

11+√2+1√2+√3+1√3+√4+........+1√24+√25 =..............

The correct option is D 4→ General term, Tr=1√r+1+√r⇒Tr=√r+1−√r(r+1)−r=√r+1−√r⇒S=∑Tr=√2−√1+√3−√2=√25−√1√4−√3=5−1=4√25−√24⇒(D)

$\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + . . . . . . . . + \frac{1}{\sqrt{24} + \sqrt{25}}$ =..............