IBDP MAI : AHL 1.10 Simplifying expressions AI HL Paper 3
▶️ Answer/Explanation
a
(i) Method 1: Exponent Rules
\( v_n = 2^{u_n} \), \( v_{n+1} = 2^{u_{n+1}} \), \( u_{n+1} = u_n + d \)
\( \frac{v_{n+1}}{v_n} = \frac{2^{u_{n+1}}}{2^{u_n}} = 2^{u_{n+1} – u_n} = 2^d \)
Method 2: Explicit Formula
\( u_n = a + (n-1)d \), so \( v_n = 2^{a + (n-1)d} \), \( v_{n+1} = 2^{a + nd} \)
\( \frac{v_{n+1}}{v_n} = \frac{2^{a + nd}}{2^{a + (n-1)d}} = 2^d \)
\( 2^d \) is constant.
(ii) For \( n=1 \): \( v_1 = 2^{u_1} = 2^a \)
(iii) \( v_n = 2^{u_n} = 2^{a + (n-1)d} \)
Result: (i) \( \frac{v_{n+1}}{v_n} = 2^d \), (ii) \( 2^a \), (iii) \( v_n = 2^{a + (n-1)d} \) [4].
b
(i) \( \{ v_n \} \) is geometric with first term \( 2^a \), ratio \( 2^d \)
\( S_n = \frac{2^a ( (2^d)^n – 1 )}{2^d – 1} = \frac{2^a (2^{dn} – 1)}{2^d – 1} \)
(ii) Method 1: Convergence Condition
Sum exists if \( |2^d| < 1 \implies 2^d < 1 \implies d \ln 2 < 0 \implies d < 0 \)
Method 2: Exponent Behavior
If \( d > 0 \), \( 2^d > 1 \), sum diverges; if \( d < 0 \), \( 2^d < 1 \), sum converges
(iii) \( S_\infty = \frac{2^a}{1 – 2^d} \)
(iv) \( \frac{2^a}{1 – 2^d} = 2^{a+1} \implies \frac{1}{1 – 2^d} = 2 \)
\( 1 = 2 – 2^{d+1} \implies 2^{d+1} = 1 \implies d = -1 \)
Result: (i) \( \frac{2^a (2^{dn} – 1)}{2^d – 1} \), (ii) \( d < 0 \), (iii) \( \frac{2^a}{1 – 2^d} \), (iv) \( d = -1 \) [8].
c
Method 1: Arithmetic Sequence Sum
\( w_n = p q^{n-1} \), \( z_n = \ln (p q^{n-1}) = \ln p + (n-1) \ln q \)
\( \{ z_n \} \) is arithmetic with first term \( \ln p \), difference \( \ln q \)
\( \sum_{i=1}^n z_i = \frac{n}{2} [2 \ln p + (n-1) \ln q] = n \ln \left( p q^{\frac{n-1}{2}} \right) = \ln \left( p^n q^{\frac{n(n-1)}{2}} \right) \)
Method 2: Direct Log Sum
\( \sum_{i=1}^n z_i = \ln (p \cdot p q \cdot p q^2 \cdots p q^{n-1}) \)
\( = \ln \left( p^n q^{0 + 1 + \cdots + (n-1)} \right) = \ln \left( p^n q^{\frac{n(n-1)}{2}} \right) \)
Result: \( \ln \left( p^n q^{\frac{n(n-1)}{2}} \right) \) [6].