Question

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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

Answer 1

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:

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

Then prove that
`S_(k+1)=` 1/2 - 1/((k + 2) + 1):

• 
  `S_(k+1)=S_k+a_(k+1)=`

• 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.

Answer 2

`S_n=(1)/(2)-(1)/(n+2)`

`S_1=(1)/(2 \cdot 3)`

`S_k=(1)/(2)-(1)/(k+2)`

`a_(k+1)&=(1)/((k+2)(k+3))`

`S_(k+1)=(1)/(2)-(1)/(k+3)`

Explanation:

Given sequence:

`(1)/((2\cdot3))+(1)/((3\cdot4))+(1)/((4\cdot5))+(1)/((5\cdot6))+...+(1)/((n+1)(n+2))`

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

`=(3-2)/((2\cdot3))+(4-3)/((3\cdot4))+(5-4)/((4\cdot5))+(6-5)/((5\cdot6))+...+((n+2)-(n+1))/((n+1)(n+2))`

Simplify:

`=\left((3)/(6)-(2)/(6)\right)+\left((4)/(12)-(3)/(12)\right)+\left((5)/(20)-(4)/(20)\right)+\left((6)/(30)-(5)/(30)\right)+...+\left((1)/((n+1))-(1)/((n+2))\right)`

`=\left((1)/(2)-(1)/(3)\right)+\left((1)/(3)-(1)/(4)\right)+\left((1)/(4)-(1)/(5)\right)+\left((1)/(5)-(1)/(6)\right)+...+\left((1)/((n+1))-(1)/((n+2))\right)`

All fractions cancel except the first and last.

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

`\boxed{S_n=(1)/(2)-(1)/(n+2)}`

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

`\begin{aligned}n=1 \implies S_1&=(1)/(2)-(1)/(1+2)\\\\&=(1)/(2)-(1)/(3)\\\\&= (3)/(2 \cdot3)-(2)/(3 \cdot2)\\\\&= (3-2)/(2 \cdot3)\\\\&=(1)/(2 \cdot3) \end{aligned}`

Hence proving the validity of the formula.

Assume the formula is valid for n = k :

`\implies S_k=(1)/((2\cdot3))+(1)/((3\cdot4))+(1)/((4\cdot5))+(1)/((5\cdot6))+...+(1)/((k+1)(k+2))`

`\implies S_k=(1)/(2)-(1)/(k+2)`

Therefore:

`\begin{aligned}S_(k+1)&=(1)/(2)-(1)/(k+1+2)}\\\\&= (1)/(2)-(1)/(k+3)}\end{aligned}`

`\textsf{Given} \; \; \; S_(k+1)=S_k+a_(k+1)\;\;\; \textsf{then}:`

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

`\textsf{Use the equation for\;\;$a_(k+1)$\;\;and\;\;$S_k$\;\;to find the equation for\;\;$S_(k+1)$}:`

`\implies S_(k+1)=S_k+a_(k+1)`

`\implies S_(k+1)=(1)/(2)-(1)/(k+2)}+(1)/((k+2)(k+3))`

`\implies S_(k+1)=(1)/(2)+(1)/((k+2)(k+3))-(1)/(k+2)}`

`\implies S_(k+1)=(1)/(2)+(1)/((k+2)(k+3))-(k+3)/((k+2)(k+3))`

`\implies S_(k+1)=(1)/(2)-(k+2)/((k+2)(k+3))`

`\implies S_(k+1)=(1)/(2)-(1)/(k+3)`