If a1=1 and an+1=ann+2 for n≥1, determine a1,a2,a3, and a4

Answer:

a1 =1

a2= 1/3

a3= 1/12

a4 = 1/60

Step-by-step explanation:

a(n+1) = an/(n+2)

if n=1

a(1+1)= a1/(1+2)

a2= a1/3

a2= 1/3

if n=2

a(2+1)= a2/(2+2)

a3= a2/4 = a2 × 1/4

a3= 1/3 × 1/4 = 1/12

if n=3

a(3+1)= a3/(3+2)

a4= a3/5 = a3 × 1/5

a4= 1/12 × 1/5 = 1/60

The value of [tex]\rm a_{1}, \rm a_{2}, \rm a_{3}, \rm a_{4}[/tex] for a given expression [tex]\rm a_{n+1} = {\frac {a_{n}} {n+2}[/tex] is 1, 1/3, 1/12, and 1/60.

We have given [tex]\rm a_{1} =1[/tex] and [tex]\rm a_{n+1} = {\frac {a_{n}} {n+2}[/tex] for n≥1 we need to determine the value of [tex]\rm a_{1}, \rm a_{2}, \rm a_{3}, \rm a_{4}[/tex] .

To solve for n terms, we will substitute the value of n = 1,2,3,4 and so on sequence-wise in the given expression.

if n = 1

[tex]\rm a_{1+1} = {\frac {a_{1}} {1+2}[/tex]

[tex]\rm a_{2} = {\frac {a_{1}} {3}}\\\rm a_{2} = {\frac{1} {3}[/tex]

if n = 2

[tex]\rm a_{2+1} = {\frac {a_{2}} {2+2}}\\\rm a_{3} = {\frac {1/3} {4}}\\\rm a_{3} = 1/12[/tex]

if n = 3

[tex]\rm a_{3+1} = {\frac {a_{3}} {3+2}}\\\rm a_{4} = {\frac {a_{3}} {5}}\\\rm a_{4} = 1/12 \times 1/5\\\rm a_{4} = 1/60[/tex]

So, the value of [tex]\rm a_{1}, \rm a_{2}, \rm a_{3}, \rm a_{4}[/tex] is 1, 1/3, 1/12, and 1/60.

Learn more about the substitution of values here:

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