If $${a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}$$, then $${a_1} + {a_2}\, + \,....\, + \,{a_{25}}$$ is equal to :

For three positive integers p, q, r, $${x^{p{q^2}}} = {y^{qr}} = {z^{{p^2}r}}$$ and r = pq + 1 such that 3, 3 log$$_yx$$, 3 log$$_zy$$, 7 log$$_xz$$ are in A.P. with common difference $$\frac{1}{2}$$. Then r-p-q is equal to

$$ \begin{aligned} &\text { Let }\left\{a_{n}\right\}_{n=0}^{\infty} \text { be a sequence such that } a_{0}=a_{1}=0 \text { and } \\\\ &a_{n+2}=3 a_{n+1}-2 a_{n}+1, \forall n \geq 0 . \end{aligned} $$

Then $$a_{25} a_{23}-2 a_{25} a_{22}-2 a_{23} a_{24}+4 a_{22} a_{24}$$ is equal to

Consider the sequence $$a_{1}, a_{2}, a_{3}, \ldots$$ such that $$a_{1}=1, a_{2}=2$$ and $$a_{n+2}=\frac{2}{a_{n+1}}+a_{n}$$ for $$\mathrm{n}=1,2,3, \ldots .$$ If $$\left(\frac{\mathrm{a}_{1}+\frac{1}{\mathrm{a}_{2}}}{\mathrm{a}_{3}}\right) \cdot\left(\frac{\mathrm{a}_{2}+\frac{1}{\mathrm{a}_{3}}}{\mathrm{a}_{4}}\right) \cdot\left(\frac{\mathrm{a}_{3}+\frac{1}{\mathrm{a}_{4}}}{\mathrm{a}_{5}}\right) \ldots\left(\frac{\mathrm{a}_{30}+\frac{1}{\mathrm{a}_{31}}}{\mathrm{a}_{32}}\right)=2^{\alpha}\left({ }^{61} \mathrm{C}_{31}\right)$$, then $$\alpha$$ is equal to :