Use mathematical induction to prove the following for every positive integer n. $$\sum _ { i = 1 } ^ { n } \frac { 17 } { i ( i + 1 ) ( i + 2 ) } = \frac { 17 n ( n + 3 ) } { 4 ( n + 1 ) ( n + 2 ) }$$

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The Answer of Use mathematical induction to prove the following for every positive integer n. $$\sum _ { i = 1 } ^ { n } \frac { 17 } { i ( i + 1 ) ( i + 2 ) } = \frac { 17 n ( n + 3 ) } { 4 ( n + 1 ) ( n + 2 ) }$$

Use Mathematical Induction to Prove the Following for Every Positive ∑i=1n17i(i+1)(i+2)=17n(n+3)4(n+1)(n+2)\sum _ { i = 1 } ^ { n } \frac { 17 } { i ( i + 1 ) ( i + 2 ) } = \frac { 17 n ( n + 3 ) } { 4 ( n + 1 ) ( n + 2 ) }i=1∑n​i(i+1)(i+2)17​=4(n+1)(n+2)17n(n+3)​

Use Mathematical Induction to Prove the Following for Every Positive $\sum _ { i = 1 } ^ { n } \frac { 17 } { i ( i + 1 ) ( i + 2 ) } = \frac { 17 n ( n + 3 ) } { 4 ( n + 1 ) ( n + 2 ) }$