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Algebra II | Progressions

It’d be very much appreciated if someone could a.) solve this for me, and b.) kindfully explain how to get the correct answer in simple terms. It looks easy, but the lesson isn’t making it make sense. Thank you! :)

Algebra II Progressions Itd be very much appreciated if someone could a solve this for me and b kindfully explain how to get the correct answer in simple terms class=

Respuesta :

devishri1977

Answer:

Option 1

Step-by-step explanation:

Substiute the value of n with 1 to 6 and find each term.

When n = 1 ,

       [tex]\sf First \ term = \dfrac{1}{2*1+1}=\dfrac{1}{2+1}=\dfrac{1}{3}\\\\[/tex]

[tex]\sf When n = 2,\\\\Second \ term = \dfrac{2}{2*2+1}=\dfrac{2}{4+1}=\dfrac{2}{5}\\\\[/tex]

When n = 3 ,

      [tex]\sf Third \ term = \dfrac{3}{2*3+1}=\dfrac{3}{6+1}=\dfrac{3}{7}[/tex]

When n = 4,

    [tex]\sf Fourth \ term =\dfrac{4}{2*4+1}=\dfrac{4}{8+1}=\dfrac{4}{9}\\[/tex]

When n = 5 ,

   [tex]\sf Fifth \ term = \dfrac{5}{2*5+1}=\dfrac{5}{10+1}=\dfrac{5}{11}\\\\When \ n = 6,\\\\ \: Sixth \ term =\dfrac{6}{6*2+1}=\dfrac{6}{12+1}=\dfrac{6}{13}[/tex]

[tex]Answer: \dfrac{1}{3}+\dfrac{2}{5}+\dfrac{3}{7}+\dfrac{4}{9}+\dfrac{5}{11}+\dfrac{6}{13}[/tex]