Mathematics - Sequence & Series Question with Solution | TestHub
Match each item in List I to all applicable entries in List II.
List I | List II |
|---|---|
(A) Given F(1)=2 and F(n+1)=(2F(n)+1)/2, find F(101). | (P) 42 |
(B) For a₁,...,a₂₁ in A.P., if a₃+a₅+a₁₁+a₁₇+a₁₉=10, find Σ(i=1 to 21) aᵢ. | (Q) 1620 |
(C) For the sequence 1,5,13,29,... defined by t₁=1 and tₙ₊₁=2tₙ+3, find its tenth term. | (R) 52 |
(D) Find the sum of two-digit positive integers divisible by neither 2 nor 3. | (S) 2045 |
| (T) 2+4+6+...+12 |
Options:
Answer:
Solution:
(A) Each recurrence step adds 1/2, so F(101)=2+100/2=52. (B) The five indices sum to 55, so the five terms sum to 5a₁₁=10. Hence a₁₁=2 and the full sum is 21a₁₁=42, also 2+4+6+8+10+12. (C) With t₁=1 and tₙ₊₁=2tₙ+3, induction gives tₙ=2ⁿ⁺¹-3, so t₁₀=2045. (D) Inclusion-exclusion gives Σ(10 to 99)k-Σ(multiples of 2)k-Σ(multiples of 3)k+Σ(multiples of 6)k=4905-2430-1665+810=1620. Thus A-R; B-P,T; C-S; D-Q.
