meredith48034
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Find a formula for the sum for the first n terms of the sequence.

1/(2*3) + 1/(3*4) + 1/(4*5) + 1/(5*6), . . ., 1/(n+1)(n+2), . . .
S_n = _______
Prove the validity of your formula.
Verify the formula for n = 1
S_1 = __________

Assume that the formula is valid for n = k.

S_k = 1/(2*3) + 1/(3*4) + 1/(4*5) + 1/(5*6) + . . . + 1/(k+1)(k+2) = _________

Then, S_(k+1) = S_k + a_(k+1) = 1/(2*3) + 1/(3*4) + 1/(4*5) + 1/(5*6) + . . . + 1/(k+1)(k+2) + a_(k+1).

a_(k+1) = __________

Use the equation for a_(k+1) and S_k to find the equation for S_(k+1).
S_(k+1)= _____________

Thus, the formula is valid

Respuesta :

semsee45

Answer:

[tex]S_n=\dfrac{1}{2}-\dfrac{1}{n+2}[/tex]

[tex]S_1=\dfrac{1}{2 \cdot 3}[/tex]

[tex]S_k=\dfrac{1}{2}-\dfrac{1}{k+2}[/tex]

[tex]a_{k+1}&=\dfrac{1}{(k+2)(k+3)}[/tex]

[tex]S_{k+1}=\dfrac{1}{2}-\dfrac{1}{k+3}[/tex]

Step-by-step explanation:

Given sequence:

[tex]\dfrac{1}{(2\cdot3)}+\dfrac{1}{(3\cdot4)}+\dfrac{1}{(4\cdot5)}+\dfrac{1}{(5\cdot6)}+...+\dfrac{1}{(n+1)(n+2)}[/tex]

Rewrite the numerator as the subtraction of the first number of the denominator from the second number of the denominator:

[tex]=\dfrac{3-2}{(2\cdot3)}+\dfrac{4-3}{(3\cdot4)}+\dfrac{5-4}{(4\cdot5)}+\dfrac{6-5}{(5\cdot6)}+...+\dfrac{(n+2)-(n+1)}{(n+1)(n+2)}[/tex]

Simplify:

[tex]=\left(\dfrac{3}{6}-\dfrac{2}{6}\right)+\left(\dfrac{4}{12}-\dfrac{3}{12}\right)+\left(\dfrac{5}{20}-\dfrac{4}{20}\right)+\left(\dfrac{6}{30}-\dfrac{5}{30}\right)+...+\left(\dfrac{1}{(n+1)}-\dfrac{1}{(n+2)}\right)[/tex]

[tex]=\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+\left(\dfrac{1}{4}-\dfrac{1}{5}\right)+\left(\dfrac{1}{5}-\dfrac{1}{6}\right)+...+\left(\dfrac{1}{(n+1)}-\dfrac{1}{(n+2)}\right)[/tex]

All fractions cancel except the first and last.  

Therefore the formula for the sum of the given sequence is:

[tex]\boxed{S_n=\dfrac{1}{2}-\dfrac{1}{n+2}}[/tex]

Substitute n = 1 into the formula to prove its validity:

[tex]\begin{aligned}n=1 \implies S_1&=\dfrac{1}{2}-\dfrac{1}{1+2}\\\\&=\dfrac{1}{2}-\dfrac{1}{3}\\\\&= \dfrac{3}{2 \cdot3}-\dfrac{2}{3 \cdot2}\\\\&= \dfrac{3-2}{2 \cdot3}\\\\&=\dfrac{1}{2 \cdot3} \end{aligned}[/tex]

Hence proving the validity of the formula.

Assume the formula is valid for n = k :

[tex]\implies S_k=\dfrac{1}{(2\cdot3)}+\dfrac{1}{(3\cdot4)}+\dfrac{1}{(4\cdot5)}+\dfrac{1}{(5\cdot6)}+...+\dfrac{1}{(k+1)(k+2)}[/tex]

[tex]\implies S_k=\dfrac{1}{2}-\dfrac{1}{k+2}[/tex]

Therefore:

[tex]\begin{aligned}S_{k+1}&=\dfrac{1}{2}-\dfrac{1}{k+1+2}}\\\\&= \dfrac{1}{2}-\dfrac{1}{k+3}}\end{aligned}[/tex]

[tex]\textsf{Given} \; \; \; S_{k+1}=S_k+a_{k+1}\;\;\; \textsf{then}:[/tex]

[tex]\begin{aligned}S_{k+1}&=S_k+a_{k+1}\\\\\dfrac{1}{2}-\dfrac{1}{k+3}&=\dfrac{1}{2}-\dfrac{1}{k+2}+a_{k+1}\\\\a_{k+1}&=\dfrac{1}{2}-\dfrac{1}{k+3}-\left(\dfrac{1}{2}-\dfrac{1}{k+2}\right)\\\\a_{k+1}&=-\dfrac{1}{k+3}+\dfrac{1}{k+2}\\\\a_{k+1}&=\dfrac{1}{k+2}-\dfrac{1}{k+3}\\\\a_{k+1}&=\dfrac{k+3}{(k+2)(k+3)}-\dfrac{k+2}{(k+2)(k+3)}\\\\a_{k+1}&=\dfrac{k+3-(k+2)}{(k+2)(k+3)}\\\\a_{k+1}&=\dfrac{1}{(k+2)(k+3)}\end{aligned}[/tex]

[tex]\textsf{Use the equation for\;\;$a_{k+1}$\;\;and\;\;$S_k$\;\;to find the equation for\;\;$S_{k+1}$}:[/tex]

[tex]\implies S_{k+1}=S_k+a_{k+1}[/tex]

[tex]\implies S_{k+1}=\dfrac{1}{2}-\dfrac{1}{k+2}}+\dfrac{1}{(k+2)(k+3)}[/tex]

[tex]\implies S_{k+1}=\dfrac{1}{2}+\dfrac{1}{(k+2)(k+3)}-\dfrac{1}{k+2}}[/tex]

[tex]\implies S_{k+1}=\dfrac{1}{2}+\dfrac{1}{(k+2)(k+3)}-\dfrac{k+3}{(k+2)(k+3)}[/tex]

[tex]\implies S_{k+1}=\dfrac{1}{2}-\dfrac{k+2}{(k+2)(k+3)}[/tex]

[tex]\implies S_{k+1}=\dfrac{1}{2}-\dfrac{1}{k+3}[/tex]

mhanifa

Given sequence with each term 1/(n + 1)(n+2).

First rewrite each term as:

  • 1/(n + 1)(n + 2) = [(n + 2) - (n - 1)] / (n + 1)(n + 2) = 1/(n + 1) - 1/(n+2)

Find the sum:

  • S ₙ =
  • 1/(2*3) + 1/(3*4) + 1/(4*5) + ... + 1/(n + 1)(n + 2) =
  • 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/(n + 1) - 1/(n + 2) =
  • 1/2 - 1/(n + 2)

Verify the formula for S₁:

  • S₁ = 1/2 - 1/3 = 3/6 - 2/6 = 1/6 = 1/(2*3)

Assume that the sum is valid for n = k:

  • [tex]S_k=[/tex] 1/(2*3) + 1/(3*4) + 1/(4*5) + ... + 1/(k + 1)(k + 2) = 1/2 - 1(k + 2)

Then prove that [tex]S_{k+1}=[/tex] 1/2 - 1/((k + 2) + 1):

  • [tex]S_{k+1}=S_k+a_{k+1}=[/tex]
  • 1/(2*3) + 1/(3*4) + 1/(4*5) + ... + 1/(k + 1)(k + 2) + 1/(k + 2)(k + 3) =
  • 1/2 - 1/(k + 2) + 1/(k + 2) - 1/(k + 3) =
  • 1/2 - 1/(k + 3) =
  • 1/2 - 1/((k + 2) + 1)

Thus, the formula is valid.

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