Use mathematical induction to solve for all positive integers n. $16 + 34 + 52 + 70 + \ldots + ( 18 n - 2 ) = ?$
A) $\frac { n ( n + 16 ) ( 16 n + 1 ) } { 6 }$
B) $\frac { n } { 2 } ( 3 n + 1 )$
C) $\frac { n } { 2 } ( 3 n + 14 )$
D) $n ( n + 16 )$
E) $\frac { n ( n + 16 ) } { 2 }$

Question

Use mathematical induction to solve for all positive integers n.​ $16 + 34 + 52 + 70 + \ldots + ( 18 n - 2 ) = ?$ ​
A) $\frac { n ( n + 16 ) ( 16 n + 1 ) } { 6 }$
B) $\frac { n } { 2 } ( 3 n + 1 )$
C) $\frac { n } { 2 } ( 3 n + 14 )$
D) $n ( n + 16 )$
E) $\frac { n ( n + 16 ) } { 2 }$

Quizplus · Accepted Answer

The given series is an arithmetic sequence with the first term $a = 16$ and common difference $d = 18$. The nth term of an arithmetic sequence is given by $a_n = a + (n - 1)d$. Substituting $a = 16$ and $d = 18$, we get $a_n = 16 + (n - 1)18 = 18n - 2$, which matches the given nth term in the problem. The sum of the first n terms of an arithmetic sequence is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$. Substituting $a = 16$, $d = 18$, and simplifying, we get $S_n = \frac{n}{2}[32 + (n - 1)18] = \frac{n}{2}[32 + 18n - 18] = \frac{n}{2}(14 + 18n) = \frac{n}{2}(3n + 14)$, which matches choice C.

Use Mathematical Induction to Solve for All Positive Integers N 16+34+52+70+…+(18n−2)=?16 + 34 + 52 + 70 + \ldots + ( 18 n - 2 ) = ?16+34+52+70+…+(18n−2)=?

Use Mathematical Induction to Solve for All Positive Integers N $16 + 34 + 52 + 70 + \ldots + ( 18 n - 2 ) = ?$