Algebra: Sequence and Series Test-2

Let p = Sum of cubes of all natural numbers till 10
and q = sum of cubes of all natural numbers till 4 The given expression=p-q
$ \begin{array}{l}{{\left( \frac{10(10+1)}{2} \right)}^{2}}-{{\left( \frac{4(4+1)}{2} \right)}^{2}}\\={{55}^{2}}-{{10}^{2}}\\=65X45\\=2925\end{array}$

The general term is square of previous term +1.
Thus the next term is 26²+1= 677.

The n^th term is n²-1.
Thus the 6^th term is 6²-1=35.

$ \displaystyle \begin{array}{l}{{2}^{3}}+{{4}^{3}}+{{6}^{3}}+.......+{{20}^{3}}\\={{\left( 2\times 1 \right)}^{3}}+{{\left( 2\times 2 \right)}^{3}}+{{\left( 2\times 3 \right)}^{3}}+.......+{{\left( 2\times 10 \right)}^{3}}\\=8\times {{1}^{3}}+8\times {{2}^{3}}+8\times {{3}^{3}}.......+8\times {{10}^{3}}\\=8\times \left[ {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+........+{{10}^{3}} \right]\\=8\times 3025=24200\\\left[ \because \,\,\,{{1}^{3}}+{{2}^{3}}+{{3}^{3}}+.........+{{10}^{3}}=3025\,\,\left( given \right) \right]\end{array}$

The sequence 1,-1, 2,-2 repeats.
Thus we just need to find the (507 mod 4) = 3^rd term of the sequence.
Thus the 507^th term will be 2.