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18 Ιαν 2020 · Learn how to prove a series formula using mathematical induction in this easy-to-follow video with examples and solutions.
11 Μαρ 2024 · The sum of the arithmetic progression 1+5+9+− (4n-3)=n (2n-1) 1 + 5 + 9 + −(4n − 3) = n(2n− 1) for integral values of n>0 n > 0. 1 Identify the first term of the arithmetic progression (AP), which is a1=1 a1 = 1. 2 Identify the nth term of the AP, which is an =4n-3 =4n−3.
25 Αυγ 2018 · Best answer. Let P (n): 1 + 5 + 9 + … + (4n – 3) = n (2n – 1), for all natural numbers n. P (1): 1 = 1 (2 x 1 – 1) = 1, which is true. Hence, P (1) is true. Let us assume that P (n) is true for some natural number n = k. ∴ P (k) :1 + 5 + 9 +…+ (4k -3) = k (2k - 1) ....... (i) Now, we have to prove that P (k + 1) is true.
18 Απρ 2017 · Explanation: Having proved, by Mathematical Induction, the Result : P (n):1 + 5 +9 +... + (4n −3) = n(2n − 1). To verify P (n) for n = 3, on the L.H.S., we need consider only. first 3 terms. ∴ The L.H.S. of P (n) (for n = 3) = 1 +5 +9 = 15...(1). The R.H.S. of P (n) (for n = 3) = 3(2 ⋅ 3 − 1) = 3 ⋅ 5 = 15...(2). ∴ The L.H.S.=The R.H.S.
28 Μαΐ 2016 · Using mathematical induction, I have proved that $$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$ for every integer $n > 0$. I would like to know if there is another way of proving this result wi...
Find step-by-step Algebra 2 solutions and your answer to the following textbook question: Prove that 1 + 5 + 9 + ... + (4n - 3) = n(2n - 1) for every natural number n..
21 Σεπ 2020 · Best answer. Let P (n) : 1 + 5 + 9 + … + (4n – 3) = n (2n – 1), ∀ n ∈ N. Step 1: P (1) : 1 = 1 (2.1 – 1) = 1 which is true for P (1) Step 2: P (k) : 1 + 5 + 9 + … + (4k – 3) = k (2k – 1). Let it be true. Step 3: P (k + 1) : 1 + 5 + 9 + … + (4k – 3) = k (4k + 1) = k (2k – 1) + (4k + 1) = 2k2 – k + 4k + 1. = 2k2 + 3k + 1 = 2k2 + 2k + k + 1.
