$The distributive law from algebra states that for all real numbers c, a_{1} and a_{2}, we have c (a_{1} + a_{2}) = c a_{1} + c a_{2} . Use this law and mathematical induction to prove that, for all natural numbers, n \geq 2, if c, a_{1}, a_{2}, . . ., a_{n} are any real numbers, then c (a_{1} + a_{2} + . . . + a_{n}) = c a_{1} + c a_{2} + . . . + c a_{n}$

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$Given : For all real numbers c, a_{1} and a_{2}, c (a_{1} + a_{2}) = c a_{1} + c a_{2} . To prove : For all natural numbers, n \geq 2, i f c, a_{1}, a_{2}, . . ., a_{n} are any real numbers, then c (a_{1} + a_{2} + . . . + a_{n}) = c a_{1} + c a_{2} + . . . + c a_{n} Proof : Let P (n) : c (a_{1} + a_{2} + . . . + a_{n}) = c a_{1} + c a_{2} + . . . + c a_{n} for all natural numbers n \geq 2 and c, a_{1}, a_{2}, . . ., a_{n} \in R . Step I : For n = 2, P (2) : LHS = c (a_{1} + a_{2}) RHS = c a_{1} + c a_{2} As, c (a_{1} + a_{2}) = c a_{1} + c a_{2} (Given) \Rightarrow LHS = RHS So, it is true for n = 2 . Step II : For n = k, Let P (k) : c (a_{1} + a_{2} + . . . + a_{k}) = c a_{1} + c a_{2} + . . . + c a_{k} be true for some natural numbers k \geq 2 and c, a_{1}, a_{2}, . . ., a_{k} \in R . Step III : For n = k + 1, P (k + 1) : LHS = c (a_{1} + a_{2} + . . . + a_{k} + a_{k + 1}) = c [(a_{1} + a_{2} + . . . + a_{k}) + a_{k + 1}] = c (a_{1} + a_{2} + . . . + a_{k}) + c a_{k + 1} = c a_{1} + c a_{2} + . . . + c a_{k} + c a_{k + 1} (Using step II) RHS = c a_{1} + c a_{2} + . . . + c a_{k} + c a_{k + 1} As, LHS = RHS So, it is also true for n = k + 1 . Hence, for all natural numbers, n \geq 2, i f c, a_{1}, a_{2}, . . ., a_{n} are any real numbers, then c (a_{1} + a_{2} + . . . + a_{n}) = c a_{1} + c a_{2} + . . . + c a_{n} .$

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The distributive law from algebra states that for all real numbers, c, $a_{1} a n d a_{2}$ , we have $c (a_{1} + a_{2}) = c a_{1} + c a_{2}$ .

Use this law and mathematical induction to prove that, for all natural numbers, $n \geq 2$ , if c, $a_{1}, a_{2}, . . . . . a_{n}$ are any numbers, then

$c (a_{1} + a_{2} + . . . . + a_{n}) = c a_{1} + c a_{2} + . . . . . . + c a_{n}$ .

Q. Let

a_{1} = 0

and

a_{1}

a_{2}

a_{3}

, ...,

a_{n}

be real numbers such that

| a_{i} | = | a_{i - 1} + 1 |

for all i then the AM of the numbers

a_{1}

a_{2}

a_{3}

, ...,

a_{n}

has the value A where

Q. If

a_{1}, a_{2}, . . . . . . ., a_{n} (n > 3)

are all unequal positive real numbers, and

E = \frac{(1 + a_{1} + a_{1}^{2}) (1 + a_{2} + a_{2}^{2}) . . . . . . (1 + a_{n} + a_{n}^{2})}{a_{1}, a_{2}, . . . . . . ., a_{n}}

then which of the following best describes E?

Q. Let a, b, c be positive real numbers. The sequence

(a_{n}) n \geq 1

is defined by

a_{1} = a, a_{2} = b

and

a_{n + 1} = \frac{a_{n}^{2} + c}{a_{n - 1}}

for all

n \geq 2.

Prove that the terms of the sequence are all positive integers if and only if a,b and

\frac{a^{2} + b^{2} + c}{a b}

are positive integers.

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in arithmetic progression, where $a_{1} > 0$ for all i.
Prove that $\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \dots + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} = \frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}$

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The distributive law from algebra states that for all real numbers c, a1 and a2, we have ca1+a2=ca1+ca2.Use this law and mathematical induction to prove that, for all natural numbers, n≥2, if c, a1, a2, ..., an are any real numbers, thenca1+a2+...+an=ca1+ca2+...+can