Summation with the form:
$$\sum_{b=0}^a\sum_{c=0}^b c$$
I am not aware of any rule about chaining sums and getting a value in terms of the variable $a$.
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3$\begingroup$ Just use the formula of sum of first n natural numbers and squares of first n natural numbers $\endgroup$– MathStackexchangeIsMarvellousCommented Apr 26 at 6:25
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$\begingroup$ Solve the internal summation as @MathStackexchangeIsNotSoBad said, and then apply the outer summation after expanding the result of the first summation $\endgroup$– GwenCommented Apr 26 at 7:18
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1 Answer
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Use these two formula : $$\sum _{k=0}^{n}k={\frac {n(n+1)}{2}}$$ $$\sum_{k=0}^{n} k^{2} = \frac{2n^{3} + 3n^{2} + n}{6}$$ that you can prove by induction.
The inner sum becomes $\displaystyle \sum_{c=0}^b c = \frac {b(b+1)}{2}$ and so $\displaystyle \sum_{b=0}^a\sum_{c=0}^b c = \sum_{b=0}^a \frac {b(b+1)}{2} =$ $\displaystyle \frac{1}{2} \left( \sum_{b=0}^a b^2 + \sum_{b=0}^a b \right) = \frac{2a^{3} + 3a^{2} + a}{12} + \frac {a(a+1)}{4}$ that you can simplify further if needed.
