Define ak=(k2 + 1)k! and bk=a1+a2+a3+. ldots ldots ...+ak. Let (a100/b100)=(m/n) where m and n are relatively prime numbers. Then the value of (n - m) is:

Question

Define ak=(k2 + 1)k! and bk=a1+a2+a3+. ldots ldots ...+ak. Let (a100/b100)=(m/n) where m and n are relatively prime numbers. Then the value of (n - m) is: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

ak=(k2+1) k !=(k(k+1)-(k-1)) k !=k(k+1) !-(k-1) k ! a1=1⋅ 2!-0 a2=2⋅ 3!-1⋅ 2! a3=3⋅ 4!-2⋅ 3! vdots (ak = k (k + 1) ! - (k - 1) k !/H e n c e bk = k (k + 1) !) ∴ (ak/bk)=((k2 + 1) k !/k (k + 1) !)=((k2 + 1)/k (k + 1))=(k2 + 1/k2 + k) (a100/b100)=(10001/10100)=(m/n);(n - m)=99 Ans.

Q. Define ak​=(k2+1)k! and bk​=a1​+a2​+a3​+.……...+ak​. Let b100​a100​​=nm​ where m and n are relatively prime numbers. Then the value of (n−m) is:

ak​=(k2+1)k!=(k(k+1)−(k−1))k!=k(k+1)!−(k−1)k! a1​=1⋅2!−0 a2​=2⋅3!−1⋅2! a3​=3⋅4!−2⋅3! ⋮ Hencebk​=k(k+1)!ak​=k(k+1)!−(k−1)k!​ ∴bk​ak​​=k(k+1)!(k2+1)k!​=k(k+1)(k2+1)​=k2+kk2+1​ b100​a100​​=1010010001​=nm​;(n−m)=99Ans.

Q. Define $a_{k} = (k^{2} + 1) k!$ and $b_{k} = a_{1} + a_{2} + a_{3} + . \dots\dots ... + a_{k} .$ Let $\frac{a _{100}}{b _{100}} = \frac{m}{n}$ where $m$ and $n$ are relatively prime numbers. Then the value of $(n - m)$ is: