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+1 vote
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in Calculus by (30 points)
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1.) 1+4+7+...+(3n-2)=n/2(3n-1)

2.) 5+9+13+...+(4n+1)=n(2n+3)

3.) 2+4+6+...+(2n)=n²+n

1 Answer

+2 votes
by (24.8k points)
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Best answer

1.

To prove P(k + 1) is true, i.e., to prove

∴ P(k + 1) is true whenever P(k) is true.

∴ By the Principle of Mathematical Induction, P(n) is true for all n ∈ N.

3. Let P(n) :2 + 4 + 6+ …+2 n = n2 + n
P(l): 2 = l2 + 1 = 2, which is true
Hence, P(l) is true.
Let us assume that P(n) is true for some natural number n = k.
∴ P(k): 2 + 4 + 6 + .,.+2k = k2 + k  (i)
Now, we have to prove that P(k + 1) is true.
P(k + l):2 + 4 + 6 + 8+ …+2k+ 2 (k +1)
= k2 + k + 2(k+ 1)  [Using (i)]
= k2 + k + 2k + 2
= k2 + 2k+1+k+1
= (k + 1)2 + k+ 1
Hence, P(k + 1) is true whenever P(k) is true.
So, by the principle of mathematical induction P(n) is true for any natural number n.

by (30 points)
+2
Thanks this is a great help. Do you know what number 2 is?

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