If $$1^{2}+2^{2}+3^{2}+........+p^{2}$$ = $$\frac{p(p+1)(2p+1)}{6}$$, then $$1^{2}+3^{2}+5^{2}+........+17^{2}$$ is equal to:

Expression : $$1^{2}+3^{2}+5^{2}+........+17^{2}$$

= $$[1^{2}+2^{2}+3^{2}+4^{2}........+16^{2}+17^{2}]$$ $$-[2^2+4^2+.........+16^2]$$

= $$[1^{2}+2^{2}+3^{2}+4^{2}........+16^{2}+17^{2}]$$ $$-(2^2)[1^2+2^2+3^2.........+8^2]$$

= $$[\frac{17(17+1)+(34+1)}{6}]-[4\times\frac{8(8+1)(16+1)}{6}]$$

= $$[\frac{17(17+1)+(34+1)}{6}]-[4\times\frac{8(8+1)(16+1)}{6}]$$

= $$[51\times35]-[48\times17]$$

= $$17\times(105-48)=969$$

=> Ans - (D)