1-16 - 6-7 - 1-20 - 6-11 - 1-24 - 2-143 - - - 1-96 - 14-47 equals1-16 - 6-7 - 1-20 - 6-11 - 1-24 - 2-143 - - - 1-96 - 14-47 - -

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

Expression Formation

1/√16 + 6√7 +...

Question

$\frac{1}{\sqrt{16 + 6 \sqrt{7}}} + \frac{1}{\sqrt{20 + 6 \sqrt{11}}} + \frac{1}{\sqrt{24 + 2 \sqrt{143}}} + . . . . . . . + \frac{1}{\sqrt{96 + 14 \sqrt{47}}}$ equals

$\frac{1}{\sqrt{16 + 6 \sqrt{7}}} + \frac{1}{\sqrt{20 + 6 \sqrt{11}}} + \frac{1}{\sqrt{24 + 2 \sqrt{143}}} + . . . . . . . + \frac{1}{\sqrt{96 + 14 \sqrt{47}}}$ बराबर है

\frac{\sqrt{3} (\sqrt{17} - \sqrt{3})}{2}

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

\frac{\sqrt{7} (\sqrt{7} - 1)}{2}

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

\frac{\sqrt{11} (\sqrt{17} - \sqrt{11})}{2}

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

\frac{\sqrt{11} (\sqrt{11} - 1)}{2}

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

Open in App

Solution

Suggest Corrections

Similar questions

\frac{1}{12}

\frac{1}{15}

\frac{1}{17}

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Δ

a_{n}

a_{1} = \frac{1}{2}, a_{n} + 1 = a_{n}^{2} + a_{n}

S_{n} = \frac{1}{a_{1} + 1} + \frac{1}{a_{2} + 1} + . . . . . . . . . . . + \frac{1}{a_{n} + 1}

[S_{2012}]

{(({log}_{2} 9)^{2})}_{2}^{\frac{1}{{log}_{2} ({log}_{2} 9)}} \cdot {(\sqrt{7})}^{\frac{1}{{log}_{4} 7}}

z_{1}

z_{2}

| \frac{1}{z_{1}} + \frac{1}{z_{2}} | = | \frac{1}{z_{1}} - \frac{1}{z_{2}} |

Δ O P Q

Join BYJU'S Learning Program