Question

1/[1+1/{1+1/(1+1/2)}] in fraction form please give full explanation.............

Answer 1

Answer:

(1–1/2)=1/2

(1–1/2)(1–1/3)=1/2×2/3=1/3

(1–1/2)(1–1/3)(1–1/4)=1/2×2/3×3/4=1/3×3/4=1/4

(1–1/2)(1–1/3)(1–1/4)(1–1/5)=1/2×2/3×3/4×4/5=1/4×4/5=1/5

As you can see, the denominator of each term cancels out the numerator of the next term, so this series can be generalised to:

(1–1/2)(1–1/3)(1–1/4)…(1–1/n)=1/n

Which can also be written as:

1/2×2/3×3/4×…×((n-1)/n)=1/n

Which we could even be further generalised to:

a/b×b/c×c/d×…×m/n = a/n

Answer 2

Answer:

1/[1+1/{1+1/(1+1/2)}]

= 1/[1+1/{1+1/(3/2)}]

= 1/[1+1/{1+(2/3)}]

= 1/[1+1/{5/3}]

= 1/[1+(3/5)]

= 1/[(8/5)]

= 5/8