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If `x=2+a+a^2+oo,w h e r e|a| |
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Answer» By summing the given infinite GS, we get `x=1/((1-a))` and `y=1/((1-b))` `:. (xy)/((x+y-1))=({1/((1-a)).1/((1-b))})/({1/((1-a))+1/((1-b))-1})` `={1/((1-a)(1-b))xx((1-a)(1-b))/((1-b)+(1-a)-(1-a)(1-b))}` `=1/((1-ab))=(1-ab)^(-1)` `=(1+ab+a^(2)b^(2)+...oo)`. Hence, `(1+ab+a^(2)b^(2)+...oo)=(xy)/((x+y-1))` |
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