If un- sin n- -cos n-then show that u5-u7-u3-u5- u3-u1-

u _ n =sin^ n θ nbsp;+cos ^ n θ nbsp; Now nbsp; u _ 5 u _ 7 / u _ 3 u _ 5 nbsp;= sin^ 5 θ nbsp;+cos ^ 5 θ nbsp; sin^ 7 θ cos ^ 7 θ ...

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Second Fundamental Theorem of Calculus

If un= sin ...

Question

If $u_{n} = {sin}^{n} θ + {cos}^{n} θ,$ then show that $\frac{u_{5} - u_{7}}{u_{3} - u_{5}} = \frac{u_{3}}{u_{1}}$ .

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\frac{1}{u_{0}} - \frac{1}{u_{1}} + \frac{1}{u_{2}} - \frac{1}{u_{3}} + \dots + {(- 1)}^{n} \frac{1}{u_{n}} = \frac{1}{u_{0} +} \frac{u_{0}^{2}}{u_{1} - u_{0} +} \frac{u_{1}^{2}}{u_{2} - u_{1} +} \dots \frac{u_{n - 1}^{2}}{u_{n} - u_{n - 1}}

T_{n} = s i n^{n} θ + c o s^{n} θ

\frac{T_{3} - T_{5}}{T_{1}} = \frac{T_{5} - T_{7}}{T_{3}}

u_{1} + u_{2} + u_{3} + u_{4} + . . . . . .

lim n \to \infty \sqrt[n]{u_{n}}

< 1

> 1

If u_{n} = {cos}^{n} θ + {sin}^{n} θ, then 2 u_{6} - 3 u_{4} + 1 = 0.

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 2 & 1 & 0 3 & 2 & 1 \end{matrix} ⎤ ⎥ ⎦

U_{1}, U_{2}, U_{3}

A U_{1} = ⎡ ⎢ ⎣ \begin{matrix} 100 \end{matrix} ⎤ ⎥ ⎦, A U_{2} = ⎡ ⎢ ⎣ \begin{matrix} 230 \end{matrix} ⎤ ⎥ ⎦, A U_{3} = ⎡ ⎢ ⎣ \begin{matrix} 231 \end{matrix} ⎤ ⎥ ⎦

3 \times 3

U_{1}, U_{2}, U_{3}

| U | =

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