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Question
\[\text{ The distributive law from algebra states that for all real numbers} c, a_1 \text{ and } a_2 , \text{ we have } c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]
\[\text{ Use this law and mathematical induction to prove that, for all natural numbers, } n \geq 2, if c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n\]
Solution
\[\text{ Given: For all real numbers } c, a_1 \text{ and } a_2 , c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]
\[\text{ To prove: For all natural numbers, } n \geq 2, \text{ if } c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n \]
\[ \text{ Proof } : \]
\[\text{ Let } P\left( n \right): c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n \text{ for all natural numbers n } \geq 2 \text{ and } c, a_1 , a_2 , . . . , a_n \in R . \]
\[\text{ Step I: For } n = 2, \]
\[P\left( 2 \right): \]
\[LHS = c\left( a_1 + a_2 \right)\]
\[RHS = c a_1 + c a_2 \]
\[\text{ As, } c\left( a_1 + a_2 \right) = c a_1 + c a_2 \left( \text{ Given } \right)\]
\[ \Rightarrow LHS = RHS\]
\[\text{ So, it is true for } n = 2 . \]
\[ \text{ Step II: For } n = k, \]
\[\text{ Let } P\left( k \right): c\left( a_1 + a_2 + . . . + a_k \right) = c a_1 + c a_2 + . . . + c a_k \text{ be true for some natural numbers } k \geq 2 \text{ and } c, a_1 , a_2 , . . . , a_k \in R . \]
\[\text{ Step III: For } n = k + 1, \]
\[P\left( k + 1 \right): \]
\[LHS = c\left( a_1 + a_2 + . . . + a_k + a_{k + 1} \right)\]
\[ = c\left[ \left( a_1 + a_2 + . . . + a_k \right) + a_{k + 1} \right]\]
\[ = c\left( a_1 + a_2 + . . . + a_k \right) + c a_{k + 1} \]
\[ = c a_1 + c a_2 + . . . + c a_k + c a_{k + 1} \left( \text{ Using step } II \right)\]
\[RHS = c a_1 + c a_2 + . . . + c a_k + c a_{k + 1} \]
\[\text{ As, } LHS = RHS\]
\[\text{ So, it is also true for n } = k + 1 . \]
\[\text{ Hence, for all natural numbers,} n \geq 2, \text{ if } c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n .\]
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