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Hence, the statement holds for the $n + 1$ case. Thus by the principle of mathematical induction $1^3 + 2^3 + 3^3 + \cdots + n^3 = (1 + 2 + 3 + \cdots + n)^2$ for each $n \in \mathbb N$.
16 Απρ 2024 · Question2: Prove the following by using the principle of mathematical induction 13 + 23 + 33+ + n3 = ( ( +1)/2)^2 Let P (n) : 13 + 23 + 33 + 43 + ..+ n3 = ( ( +1)/2)^2 For n = 1, L.H.S = 13 = 1 R.H.S = (1(1 + 1)/2)^2= ((1 2)/2)^2= (1)2 = 1 Hence, L.H.S. = R.H.S P(n) is true for n = 1 Assume that P(k) is true 13 + 23 + 33 + 43 + ..+ k3 = ( ( + 1 ...
10 Φεβ 2023 · If \(\frac{1^3 + 2^3 +3^3+....\text{upto n terms}}{1.3 + 2.5 + 3.7+....\text{upto n terms}} = \frac 95\), then the value of n is ___.
Here, first term a 1 = 3-1 n. Common difference: d = a 2-a 1 = 3-2 n-3 + 1 n =-1 n. For the p th term: a p = a + (p-1) d = (3-1 n) + (p-1) (-1 n) = 3-1 n-p n + 1 n = 3-p n. Therefore, the pth term is 3-n p. Hence, the correct answer is option (B).
16 Απρ 2024 · Example 1 For all n ≥ 1, prove that 12 + 22 + 32 + 42 +…+ n2 = (n (n+1) (2n+1))/6 Let P (n) : 12 + 22 + 32 + 42 + …..+ n2 = (𝑛 (𝑛 + 1) (2𝑛 + 1))/6 Proving for n = 1 For n = 1, L.H.S = 12 = 1 R.H.S = (1 (1+1) (2 × 1+ 1))/6 = (1 × 2 × 3)/6 = 1 Since, L.H.S. = R.H.S ∴ P (n) is true for n = 1 Proving P (k + 1) is true if P (k) is true Assume tha...
27 Ιουν 2017 · Can you prove that 12 + 22 + 32 +... + n2 = 1 6 n(n + 1)(2n + 1)? see explanation. using the method of proof by induction. this involves the following steps. ∙ prove true for some value, say n = 1. ∙ assume the result is true for n = k. ∙ prove true for n = k + 1. n = 1 → LH S = 12 = 1. and RHS = 1 6 (1 + 1)(2 +1) = 1. ⇒result is true for n = 1.
6 Μαΐ 2017 · It means n-1 + 1; n-2 + 2. The result is always n. And since you are adding two numbers together, there are only (n-1)/2 pairs that can be made from (n-1) numbers.
