मराठी

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 . - Mathematics

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प्रश्न

\[\text{ The distributive law from algebra states that for all real numbers}  c, a_1 \text{ and }  a_2 , \text{ we have }  c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]
\[\text{ Use this law and mathematical induction to prove that, for all natural numbers, } n \geq 2, if c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]
\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n\]

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उत्तर

\[\text{ Given: For all real numbers } c, a_1 \text{ and }  a_2 , c\left( a_1 + a_2 \right) = c a_1 + c a_2 . \]

\[\text{ To prove: For all natural numbers, } n \geq 2, \text{ if } c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]

\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n \]

\[ \text{ Proof } : \]

\[\text{ Let } P\left( n \right): c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n \text{ for all natural numbers n } \geq 2 \text{ and  } c, a_1 , a_2 , . . . , a_n \in R . \]

\[\text{ Step I: For } n = 2, \]

\[P\left( 2 \right): \]

\[LHS = c\left( a_1 + a_2 \right)\]

\[RHS = c a_1 + c a_2 \]

\[\text{ As, }  c\left( a_1 + a_2 \right) = c a_1 + c a_2 \left( \text{ Given } \right)\]

\[ \Rightarrow LHS = RHS\]

\[\text{ So, it is true for } n = 2 . \]

\[ \text{ Step II: For } n = k, \]

\[\text{ Let }  P\left( k \right): c\left( a_1 + a_2 + . . . + a_k \right) = c a_1 + c a_2 + . . . + c a_k \text{ be true for some natural numbers }  k \geq 2 \text{ and } c, a_1 , a_2 , . . . , a_k \in R . \]

\[\text{ Step III: For }  n = k + 1, \]

\[P\left( k + 1 \right): \]

\[LHS = c\left( a_1 + a_2 + . . . + a_k + a_{k + 1} \right)\]

\[ = c\left[ \left( a_1 + a_2 + . . . + a_k \right) + a_{k + 1} \right]\]

\[ = c\left( a_1 + a_2 + . . . + a_k \right) + c a_{k + 1} \]

\[ = c a_1 + c a_2 + . . . + c a_k + c a_{k + 1} \left( \text{ Using step } II \right)\]

\[RHS = c a_1 + c a_2 + . . . + c a_k + c a_{k + 1} \]

\[\text{ As, } LHS = RHS\]

\[\text{ So, it is also true for n }  = k + 1 . \]

\[\text{ Hence, for all natural numbers,}  n \geq 2, \text{ if }  c, a_1 , a_2 , . . . , a_n \text{ are any real numbers, then } \]

\[c\left( a_1 + a_2 + . . . + a_n \right) = c a_1 + c a_2 + . . . + c a_n .\]

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पाठ 12: Mathematical Induction - Exercise 12.2 [पृष्ठ २९]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 12 Mathematical Induction
Exercise 12.2 | Q 50 | पृष्ठ २९

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