The sum 1^2-2 3^2+3 5^2-4 7^2+5 9^2-+15 29^2 is _________.

Question

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 \_\_\_\_\_\_\_\_\_.

Answer: 6952

Accepted Answer

$S={1}^{2}-{2.3}^{2}+{3.5}^{2}-{4.7}^{2}+\ldots \ldots +{15.29}^{2}$

Separating odd placed and even placed terms we get

$\begin{gathered} S=({1.1}^{2}+{3.5}^{2}+\ldots .15.(29{)}^{2})-({2.3}^{2}+{4.7}^{2} \ +\ldots .+14.(27{)}^{2} \end{gathered}$

$\begin{gathered} =\sum _{r=1}^{8}(2r-1)(4r-3{)}^{2}-\sum _{r=1}^{7}2r(4r-1{)}^{2} \ =\sum _{r=1}^{8}(32{r}^{3}-64{r}^{2}+42r-9)-2\sum _{r=1}^{7}16{r}^{3}-8{r}^{2}+r \ =32	imes {36}^{2}-64	imes 204+1512-72 \ -2(16	imes {28}^{2}-1120+28) \ =6592 \end{gathered}$

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