This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Haya Shankar, hizi hapa suluhisho za maswali yako: 4) Given the series 2 + 4 + 8 + 16 + + 16384 a) Identify the type of series Step 1: Angalia uwiano kati ya namba zinazofuatana. (4)/(2) = 2 (8)/(4) = 2 (16)/(8) = 2 Kwa kuwa uwiano ni sawa (constant ratio), hii ni geometric series. The type of series is a Geometric Series. b) Find the number of terms in the series Step 1: Tambua vigezo. First term, a = 2. Common ratio, r = 2. Last term, T_n = 16384. Step 2: Tumia fomula ya n-th term ya geometric series: T_n = a · r^n-1. 16384 = 2 · 2^n-1 16384 = 2^n Step 3: Tafuta thamani ya n. Tunajua 2^10 = 1024. 2^11 = 2048 2^12 = 4096 2^13 = 8192 2^14 = 16384 Hivyo, n = 14. The number of terms in the series is 14. c) Calculate the sum of terms in the series Step 1: Tumia fomula ya jumla ya n terms ya geometric series: S_n = (a(r^n - 1))/(r - 1) (kwa sababu r > 1). S_14 = 2(2^14 - 1)2 - 1 Step 2: Badilisha thamani na hesabu. S_14 = (2(16384 - 1))/(1) S_14 = 2(16383) S_14 = 32766 The sum of terms in the series is 32766. B) Given that the sum of the first n terms of the sequence 3 + 7 + 11 + 15 + is 820, find the last term of the sequence. Step 1: Tambua aina ya sequence na vigezo. Angalia tofauti kati ya namba zinazofuatana: 7 - 3 = 4 11 - 7 = 4 15 - 11 = 4 Kwa kuwa tofauti ni sawa (common difference), hii ni arithmetic sequence. First term, a = 3. Common difference, d = 4. Sum of n terms, S_n = 820. Step 2: Tafuta idadi ya terms, n, kwa kutumia fomula ya jumla ya arithmetic series: S_n = (n)/(2)(2a + (n-1)d). 820 = (n)/(2)(2(3) + (n-1)4) 820 = (n)/(2)(6 + 4n - 4) 820 = (n)/(2)(4n + 2) 820 = n(2n + 1) 820 = 2n^2 + n 2n^2 + n - 820 = 0 Step 3: Tatua quadratic equation kwa n kwa kutumia quadratic formula n = -b ± sqrt(b^2 - 4ac)2a. Hapa, a=2, b=1, c=-820. n = -1 ± sqrt(1^2 - 4(2)(-820))2(2) n = -1 ± sqrt(1 + 6560)4 n = -1 ± sqrt(6561)4 n = (-1 ± 81)/(4) Kuna majibu mawili: n_1 = (-1 + 81)/(4) = (80)/(4) = 20 n_2 = (-1 - 81)/(4) = (-82)/(4) = -20.5 Kwa kuwa idadi ya terms haiwezi kuwa hasi au sehemu, n = 20. Step 4: Tafuta last term (T_n) kwa kutumia fomula ya n-th term ya arithmetic sequence: T_n = a + (n-1)d. T_20 = 3 + (20-1)4 T_20 = 3 + (19)4 T_20 = 3 + 76 T_20 = 79 The last term of the sequence is 79. Last free one today — make it count tomorrow, or type /upgrade for unlimited.