Answer:
[tex]650[/tex]
Step-by-step explanation:
[tex]Well, we\ can\ solve\ the\ question\ by\ following\ some\ simple\ steps.\\Lets\ first\ set\ up\ an\ equation\ representing\ the\ Sum\ as\ S.\\Hence\ let\ this\ be\ E_1,\\S=1+2+3....50\ \\Now,\\By\ focusing\ on\ the\ RHS\ of\ the\ equation,\\1+2+3...50=50+49+48....1 [Commutative\ Property\ Of\ Addition]\\Hence\ let\ this\ be\ E_2,\\S=50+49+48....1\\By\ adding\ E_1\ and\ E_2,\\S=1+2+3...50\\S=50+49+48...1\\We\ get,\\2S=51+51+51.....51\\[/tex]
[tex]The\ 51\ in\ the\ RHS\ recurs\ 50\ times.\\Hence,\\2S=51*50\\S=\frac{51*50}{2}\\S=51*25 =1275[/tex]
[tex]Now,\\As\ we\ know\ that,\\The\ sum\ of\ all\ odd\ positive\ integers\ to\ a\ number\ n, where\ x\ is\ the\ number\ of\\ integers\ between\ 1\ and\ n, is\ given\ by\ the\ formula: \\S_{(all\ odds\ till\ n)}=x^2\\Now,\\x=\frac{(n+1)}{2}\\Hence,\\S_O= [\frac{(n+1)}{2}]^2\\Hence,\\As\ n\ can\ only\ be\ odd,\ n=50-1=49\\Hence,\\S_O=[\frac{(49+1)}{2}]^2\\S_O=25^2\\S_O=625\\If\ Sum\ of\ all\ even\ integers\ would\ be\ S_E,,\\As\ S_O+S_E=S\\Hence,\\625+S_E=1275\\Hence,\\S_E=1275-625\\S_E=650[/tex]
