Question

using mathematical induction and find 1^2^3+2^3^4+3^4^5+....upto n terms= n(n+1)(n+2)(n+3)/4

Answer 1

Answer on Question #60452 – Math – Real Analysis

Question

$$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + n(n + 1)(n + 2) = n(n + 1)(n + 2)(n + 3)/4$$

Solution

Consider the proof of (1) by mathematical induction.

1. Basis of the Induction.

Show that the statement (1) holds for $n = 1$:

the left-hand side is $1 \cdot 2 \cdot 3 = 6$,

the right-hand side is $1 \cdot (1 + 1) \cdot (1 + 2) \cdot (1 + 3) / 4 = 2 \cdot 3 \cdot 4 / 4 = 6$,

hence $6 = 6$ is true.

2. Induction hypothesis.

Assume the statement (1) holds for some $n = k$, where $k > 1$:

$$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + k(k + 1)(k + 2) = k(k + 1)(k + 2)(k + 3)/4.$$

3. Induction step.

It must then be shown that the statement (1) holds for $n = k + 1$, that is,

$$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + (k + 1)(k + 1 + 1)(k + 1 + 2) = (k + 1)(k + 1 + 1)(k + 1 + 2)(k + 1 + 3)/4;$$

i.e.

$$1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + (k + 1)(k + 2)(k + 3) = (k + 1)(k + 2)(k + 3)(k + 4)/4;$$

Consider the left-hand side of (3):

$$\begin{array}{l} 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + (k + 1)(k + 2)(k + 3) = \\ = \{\text{include the last two terms}\} = \\ = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + \dots + k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3) = \\ = \{\text{using the induction hypothesis (2) for the first } k \text{ terms}\} = \\ = \frac{k(k + 1)(k + 2)(k + 3)}{4} + (k + 1)(k + 2)(k + 3) = \\ = \{\text{factor out the common multipliers } (k + 1), (k + 2), (k + 3)\} = \\ = (k + 1)(k + 2)(k + 3) \cdot \left(\frac{k}{4} + 1\right) = (k + 1)(k + 2)(k + 3) \cdot \left(\frac{k + 4}{4}\right) = \frac{(k + 1)(k + 2)(k + 3)(k + 4)}{4}, \end{array}$$

so we deduced the right-hand side of (3).

Thus, from the assumption of the validity of formula (1) for $n = k$ it follows that it is also valid for $n = k + 1$.

4. According to the principle of mathematical induction, formula (1) is proved for all natural numbers.

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