The sum $$1^2 - 2 \cdot 3^2 + 3 \cdot 5^2 - 4 \cdot 7^2 + 5 \cdot 9^2 - \ldots + 15 \cdot 29^2$$ is ______.

The terms of the series can be written in a compact “$$n^{\text{th}}$$-term” form first and then summed with the help of the standard formulae for the alternating sums of $$n,\;n^{2},\;n^{3}$$.

Step 1 : General term
The pattern is

$$1^{2}-2\cdot 3^{2}+3\cdot 5^{2}-4\cdot 7^{2}+5\cdot 9^{2}-\ldots+15\cdot 29^{2}$$

For $$n=1,2,3,\ldots$$ an odd number $$2n-1$$ is being squared and multiplied by $$n$$, while the sign alternates beginning with $$+$$. Thus the $$n^{\text{th}}$$ term is

$$T_n \;=\;(-1)^{\,n+1}\,n\,(2n-1)^{2}\qquad (n=1,2,\ldots ,15)$$

Step 2 : Expand the algebraic factor

$$(2n-1)^{2}=4n^{2}-4n+1\;,$$ $$T_n=(-1)^{\,n+1}\,n\,(4n^{2}-4n+1) =(-1)^{\,n+1}\bigl(4n^{3}-4n^{2}+n\bigr)$$

Step 3 : Break the required sum into three separate alternating sums

$$S=\sum_{n=1}^{15}T_n =4\sum_{n=1}^{15}(-1)^{\,n+1}n^{3} -4\sum_{n=1}^{15}(-1)^{\,n+1}n^{2} +\,\sum_{n=1}^{15}(-1)^{\,n+1}n$$

Denote

$$A=\sum_{n=1}^{15}(-1)^{\,n+1}n ,\qquad B=\sum_{n=1}^{15}(-1)^{\,n+1}n^{2},\qquad C=\sum_{n=1}^{15}(-1)^{\,n+1}n^{3}.$$

Step 4 : Evaluate the three alternating sums
Because the sign is $$+$$ for odd $$n$$ and $$-$$ for even $$n$$, each of these sums is “(sum over odd) - (sum over even)”.

• Sum of the first 15 natural numbers is $$\frac{15\cdot16}{2}=120$$.
Sum over the seven even numbers $$2,4,\ldots ,14$$ is $$2(1+2+\cdots+7)=2\cdot28=56$$.
Therefore

$$A=(1+3+5+7+9+11+13+15)-56=64-56=8.$$

• Squares:

Sum of odd squares $$1^{2}+3^{2}+5^{2}+7^{2}+9^{2}+11^{2}+13^{2}+15^{2}=680,$$ sum of even squares $$2^{2}+4^{2}+6^{2}+8^{2}+10^{2}+12^{2}+14^{2}=560.$$

$$B=680-560=120.$$

• Cubes:

Sum of odd cubes $$1^{3}+3^{3}+5^{3}+7^{3}+9^{3}+11^{3}+13^{3}+15^{3}=8128,$$ sum of even cubes $$2^{3}+4^{3}+6^{3}+8^{3}+10^{3}+12^{3}+14^{3}=6272.$$

$$C=8128-6272=1856.$$

Step 5 : Assemble the result

$$\begin{aligned} S &=4C-4B+A\\ &=4(1856)-4(120)+8\\ &=7424-480+8\\ &=6952. \end{aligned}$$

Therefore the required sum equals $$6952$$.

Final Answer : 6952