Tính f(1) + f(2) + f(3) +…+ f(40)

Lời giải:

Gọi \[\sqrt {2{\rm{n}} + 1} \] = a; \[\sqrt {2{\rm{n}} - 1} \] = b

Suy ra: 4n = a² + b²; \[\sqrt {4{{\rm{n}}^2} - 1} \]= ab

Suy ra: f(n) = \[\frac{{{{\rm{a}}^3} - {{\rm{b}}^3}}}{{{{\rm{a}}^2} - {{\rm{b}}^2}}}\] = \[\frac{1}{2}.\left[ {{{\left( {\sqrt {2{\rm{n}} + 1} } \right)}^3} + {{\left( {\sqrt {2{\rm{n}} - 1} } \right)}^3}} \right]\]

 f(1) + f(2) + f(3) +…+ f(40)

= \[\frac{1}{2}.\left[ {{{\left( {\sqrt 3 } \right)}^3} - {{\left( {\sqrt 1 } \right)}^3} + {{\left( {\sqrt 5 } \right)}^3} - {{\left( {\sqrt 3 } \right)}^3} + ... + {{\left( {\sqrt {81} } \right)}^3} - {{\left( {\sqrt {79} } \right)}^3}} \right]\]

= \[\frac{1}{2}.\left[ {{{\left( {\sqrt {81} } \right)}^3} - {{\left( {\sqrt 1 } \right)}^3}} \right]\] = 364