Toán Tính E=1/2√1+1√2 + 1/3√2+2√3 +… + 1/25√24+24√25 03/10/2021 By aikhanh Tính E=1/2√1+1√2 + 1/3√2+2√3 +… + 1/25√24+24√25
Đáp án: \({4 \over 5}\) Giải thích các bước giải: $$\eqalign{ & E = {1 \over {2\sqrt 1 + 1\sqrt 2 }} + {1 \over {3\sqrt 2 + 2\sqrt 3 }} + … + {1 \over {25\sqrt {24} + 24\sqrt {25} }} \cr & {1 \over {\left( {k + 1} \right)\sqrt k + k\sqrt {k + 1} }} \cr & = {{\left( {k + 1} \right)\sqrt k – k\sqrt {k + 1} } \over {{{\left( {k + 1} \right)}^2}k – {k^2}\left( {k + 1} \right)}} \cr & = {{\left( {k + 1} \right)\sqrt k – k\sqrt {k + 1} } \over {k\left( {k + 1} \right)\left( {k + 1 – k} \right)}} \cr & = {{\left( {k + 1} \right)\sqrt k – k\sqrt {k + 1} } \over {k\left( {k + 1} \right)}} \cr & = {1 \over {\sqrt k }} – {1 \over {\sqrt {k + 1} }} \cr & \Rightarrow E = {1 \over {\sqrt 1 }} – {1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }} – {1 \over {\sqrt 3 }} + … + {1 \over {\sqrt {24} }} – {1 \over {\sqrt {25} }} \cr & \,\,\,\,\,\,E = {1 \over {\sqrt 1 }} – {1 \over {\sqrt {25} }} = 1 – {1 \over 5} = {4 \over 5} \cr} $$ Trả lời
Đáp án:
\({4 \over 5}\)
Giải thích các bước giải:
$$\eqalign{
& E = {1 \over {2\sqrt 1 + 1\sqrt 2 }} + {1 \over {3\sqrt 2 + 2\sqrt 3 }} + … + {1 \over {25\sqrt {24} + 24\sqrt {25} }} \cr
& {1 \over {\left( {k + 1} \right)\sqrt k + k\sqrt {k + 1} }} \cr
& = {{\left( {k + 1} \right)\sqrt k – k\sqrt {k + 1} } \over {{{\left( {k + 1} \right)}^2}k – {k^2}\left( {k + 1} \right)}} \cr
& = {{\left( {k + 1} \right)\sqrt k – k\sqrt {k + 1} } \over {k\left( {k + 1} \right)\left( {k + 1 – k} \right)}} \cr
& = {{\left( {k + 1} \right)\sqrt k – k\sqrt {k + 1} } \over {k\left( {k + 1} \right)}} \cr
& = {1 \over {\sqrt k }} – {1 \over {\sqrt {k + 1} }} \cr
& \Rightarrow E = {1 \over {\sqrt 1 }} – {1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }} – {1 \over {\sqrt 3 }} + … + {1 \over {\sqrt {24} }} – {1 \over {\sqrt {25} }} \cr
& \,\,\,\,\,\,E = {1 \over {\sqrt 1 }} – {1 \over {\sqrt {25} }} = 1 – {1 \over 5} = {4 \over 5} \cr} $$