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if `x=1^1/1+2^2/3+...1001^2/2001, y=1^1/3+2^2/5+...1001^2/2003` then `([x-y])/10` isA. 500B. 450C. 510D. 555 |
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Answer» Correct Answer - A::B `x-y=1^(2)((1)/(1)-(1)/(3))+2^(2)((1)/(3)-(1)/(5))….1001^(2)((1)/(2001)-(1)/(2003))` `=(1^(2).2)/(1.3)+(2^(2)2)/(3.5)…+(2.1001^(2))/(2001.2003)` `(x-y)/(2)=((1^(2))/(1.3)+(2^(2))/(3.5)…(1001^(2))/(2001.2003))` `T_(r)=(r^(2))/((2r-1)(2r+1))` `=(1)/(4)((4r^(2)+1)/((2r-1))(-1)/((2r+1)))` `=(1)/(4)(1+(1)/((2r-1)(2r+1)))` `=(1)/(4)(1)+((1)/((2r-1)-(1)/((2r+1)))` `sum_(r=1)^(1001)T_(r)=(1)/(4)(1001)+(1)/(8)(1-(1)/(2003))` `(x-y)/(2)=(1001)/(2)+(2002)/(2003.8)` `x-y=(1001)/(2)+(1001)/(2.2003)` `[x-y]=500` |
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