Use the recursive definition of summation together with mathematical induction to prove that for all positive integers $n$, if $a _ { 1 } , a _ { 2 } , \ldots , a _ { n }$ and $b _ { 1 } , b _ { 2 } , \ldots , b _ { n }$ are real numbers, then
$$\sum _ { k = 1 } ^ { n } \left( 2 a _ { k } - 3 b _ { k } \right) = 2 \sum _ { k = 1 } ^ { n } a _ { k } - 3 \sum _ { k = 1 } ^ { n } b _ { k } .$$

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The Answer of Use the recursive definition of summation together with mathematical induction to prove that for all positive integers $n$, if $a _ { 1 } , a _ { 2 } , \ldots , a _ { n }$ and $b _ { 1 } , b _ { 2 } , \ldots , b _ { n }$ are real numbers, then
$$\sum _ { k = 1 } ^ { n } \left( 2 a _ { k } - 3 b _ { k } \right) = 2 \sum _ { k = 1 } ^ { n } a _ { k } - 3 \sum _ { k = 1 } ^ { n } b _ { k } .$$

Use the Recursive Definition of Summation Together with Mathematical Induction nnn

Use the Recursive Definition of Summation Together with Mathematical Induction $n$