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If P (n) is the statement "2n ≥ 3n" and if P (r) is true, prove that P (r + 1) is true.
Concept: undefined >> undefined
If P (n) is the statement "n2 + n is even", and if P (r) is true, then P (r + 1) is true.
Concept: undefined >> undefined
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Given an example of a statement P (n) such that it is true for all n ∈ N.
Concept: undefined >> undefined
If P (n) is the statement "n2 − n + 41 is prime", prove that P (1), P (2) and P (3) are true. Prove also that P (41) is not true.
Concept: undefined >> undefined
Give an example of a statement P(n) which is true for all n ≥ 4 but P(1), P(2) and P(3) are not true. Justify your answer.
Concept: undefined >> undefined
1 + 2 + 3 + ... + n = \[\frac{n(n + 1)}{2}\] i.e. the sum of the first n natural numbers is \[\frac{n(n + 1)}{2}\] .
Concept: undefined >> undefined
12 + 22 + 32 + ... + n2 =\[\frac{n(n + 1)(2n + 1)}{6}\] .
Concept: undefined >> undefined
1 + 3 + 32 + ... + 3n−1 = \[\frac{3^n - 1}{2}\]
Concept: undefined >> undefined
\[\frac{1}{1 . 2} + \frac{1}{2 . 3} + \frac{1}{3 . 4} + . . . + \frac{1}{n(n + 1)} = \frac{n}{n + 1}\]
Concept: undefined >> undefined
1 + 3 + 5 + ... + (2n − 1) = n2 i.e., the sum of first n odd natural numbers is n2.
Concept: undefined >> undefined
\[\frac{1}{2 . 5} + \frac{1}{5 . 8} + \frac{1}{8 . 11} + . . . + \frac{1}{(3n - 1)(3n + 2)} = \frac{n}{6n + 4}\]
Concept: undefined >> undefined
\[\frac{1}{1 . 4} + \frac{1}{4 . 7} + \frac{1}{7 . 10} + . . . + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1}\]
Concept: undefined >> undefined
\[\frac{1}{3 . 5} + \frac{1}{5 . 7} + \frac{1}{7 . 9} + . . . + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)}\]
Concept: undefined >> undefined
\[\frac{1}{3 . 7} + \frac{1}{7 . 11} + \frac{1}{11 . 5} + . . . + \frac{1}{(4n - 1)(4n + 3)} = \frac{n}{3(4n + 3)}\]
Concept: undefined >> undefined
1.2 + 2.22 + 3.23 + ... + n.2n = (n − 1) 2n+1+2
Concept: undefined >> undefined
2 + 5 + 8 + 11 + ... + (3n − 1) = \[\frac{1}{2}n(3n + 1)\]
Concept: undefined >> undefined
1.3 + 2.4 + 3.5 + ... + n. (n + 2) = \[\frac{1}{6}n(n + 1)(2n + 7)\]
Concept: undefined >> undefined
1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]
Concept: undefined >> undefined
1.2 + 2.3 + 3.4 + ... + n (n + 1) = \[\frac{n(n + 1)(n + 2)}{3}\]
Concept: undefined >> undefined
Find the value of the other five trigonometric functions
Concept: undefined >> undefined
