1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2

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Solution

Let the given statement be P(n), i.e.,

P(n): 1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2

For n = 1, we have

P(1): 1.2 = 2 = (1 – 1) 2¹⁺¹ + 2 = 0 + 2 = 2, which is true.

Let P(k) be true for some positive integer k, i.e.,

1.2 + 2.2² + 3.2² + … + k.2^k = (k – 1) 2^k^{+ 1} + 2 … (i)

We shall now prove that P(k + 1) is true.

Consider

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

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Prove the following by using the principle of mathematical induction for all n ∈ N: 1.2 + 2.2² + 3.2² + … + n.2ⁿ = (n – 1) 2ⁿ⁺¹ + 2

n \in N : 1.2 + 2. 2^{2} + 3. 2^{2} + . . . . . . . + n . 2^{n} = (n - 1) 2^{n + 1} + 2

$1.2 + {2.2}^{2} + {3.2}^{3} + \dots + n . 2^{n} = (n - 1) 2^{n + 1} + 2.$

1.2 + {2.2}^{2} + {3.2}^{3} + . . . + n {.2}^{n} = (n - 1) 2^{n + 1} + 2.

1.2 + {2.2}^{2} + {3.2}^{3} + . . . . . . . . . . . . . . . . . + n {.2}^{n} = (n - 1) 2^{n + 1} + 2

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