Use mathematical induction to solve for all positive integers n. $22 + 27 + 32 + 37 + \ldots + ( 5 n - 17 ) = ?$
A) $\frac { n } { 2 } ( 5 n + 5 )$
B) $n ( 22 n + 1 )$
C) $\frac { n ( n + 1 ) } { 22 }$
D) $n ( n + 22 )$
E) $\frac { n } { 2 } ( 5 n + 1 )$

Question

Use mathematical induction to solve for all positive integers n.   $22 + 27 + 32 + 37 + \ldots + ( 5 n - 17 ) = ?$  
A) $\frac { n } { 2 } ( 5 n + 5 )$
B) $n ( 22 n + 1 )$
C) $\frac { n ( n + 1 ) } { 22 }$
D) $n ( n + 22 )$
E) $\frac { n } { 2 } ( 5 n + 1 )$

Quizplus · Accepted Answer

The given sequence is an arithmetic sequence with the first term $a_1 = 22$ and common difference $d = 5$. The nth term of an arithmetic sequence is given by $a_n = a_1 + (n - 1)d$. Substituting $a_1 = 22$ and $d = 5$, we get $a_n = 22 + (n - 1)5 = 5n + 17$. The sum of the first n terms of an arithmetic sequence is given by $S_n = \frac{n}{2}(a_1 + a_n)$. Substituting $a_1 = 22$ and $a_n = 5n + 17$, we get $S_n = \frac{n}{2}(22 + 5n + 17) = \frac{n}{2}(5n + 39) = \frac{n}{2}(5n + 5)$, which matches option A.

Use Mathematical Induction to Solve for All Positive Integers N 22+27+32+37+…+(5n−17)=?22 + 27 + 32 + 37 + \ldots + ( 5 n - 17 ) = ?22+27+32+37+…+(5n−17)=?

Use Mathematical Induction to Solve for All Positive Integers N $22 + 27 + 32 + 37 + \ldots + ( 5 n - 17 ) = ?$