Sequences and S...

Find the missing term in the series given below. $$12,\ 13,\ 18,\ 19,\ 24,\ 25 $$?

The sum of first $$9$$ terms of the series
$$\dfrac { { 1 }^{ 3 } }{ 1 } +\dfrac { { 1 }^{ 3 }+{ 2 }^{ 3 } }{ 1+3 } +\dfrac { { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 } }{ 1+3+5 } +\dots$$

Sum of the series  $$ S = 1^{2}-2^{2}+3^{2}-4^{2}+......-2002^{2}+2003^{2}$$ is

If $$\dfrac{1}{1^{2}}+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+........\infty =\dfrac{\pi^{2}}{6}$$ then $$\dfrac{1}{1^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{5^{2}}+......\infty$$

Sum of series $$\displaystyle \sum^{n}_{r=1}(r^{2}+1)_{r!}$$ is

Sum of the series 
$$1+2.2+3.2^{2}+...+100.2^{10}=$$

$$\dfrac {1}{1.4}+\dfrac {1}{4.7}+\dfrac {1}{7.10}+...+\dfrac {1}{(3n-5)(3n-2)}$$

The sum of the series $$1+\frac { 1.3 }{ 6 } +\frac { 1.3.5 }{ 6.8 } +...\infty $$ is 

Sum series $$S=1+\frac{3}{2}+\frac{5}{2^{2}}+\frac{7}{2^{3}}+....\infty $$ is 

The $${ 2006 }^{ th }$$ digit in the sequence $$12345678910111213$$.....is