If a_(1)=9 and a_(n)=na_(n-1)+1 then find the value of a_(5). Answer:

Given terms: We are given the first term of the sequence, a_(1)=9, and a recursive formula for the nth term: a_(n)=na_(n-1)+1. To find a_(5), we need to find the values of a_(2), a_(3), a_(4), and then a_(5) using the recursive formula.Find a_(2): First, let's find a_(2) using the recursive formula. We know that a_(1)=9, so: a_(2) = 2 * a_(1) + 1 = 2 * 9 + 1 = 18 + 1 = 19.Find a_(3): Next, we find a_(3) using the value of a_(2) we just found: a_(3) = 3 * a_(2) + 1 = 3 * 19 + 1 = 57 + 1 = 58.Find a_(4): Now, we find a_(4) using the value of a_(3): a_(4) = 4 * a_(3) + 1 = 4 * 58 + 1 = 232 + 1 = 233.Find a_(5): Finally, we find a_(5) using the value of a_(4): a_(5) = 5 * a_(4) + 1 = 5 * 233 + 1 = 1165 + 1 = 1166.

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Find the roots of factored polynomials

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Q. If

a_{1}=9

and

a_{n}=n a_{n-1}+1

then find the value of

a_{5}

\newline

Answer:

Given terms: We are given the first term of the sequence, $a_{1}=9$ , and a recursive formula for the $n$ th term: $a_{n}=na_{n-1}+1$ . To find $a_{5}$ , we need to find the values of $a_{2}$ , $a_{3}$ , $a_{4}$ , and then $a_{5}$ using the recursive formula.
Find $a_{2}$ : First, let's find $a_{2}$ using the recursive formula. We know that $a_{1}=9$ , so: $a_{2} = 2 \times a_{1} + 1 = 2 \times 9 + 1 = 18 + 1 = 19$ .
Find $a_{3}$ : Next, we find $a_{3}$ using the value of $a_{2}$ we just found: $\newline$ $a_{3} = 3 \times a_{2} + 1 = 3 \times 19 + 1 = 57 + 1 = 58$ .
Find $a_{4}$ : Now, we find $a_{4}$ using the value of $a_{3}$ : $a_{4} = 4 \times a_{3} + 1 = 4 \times 58 + 1 = 232 + 1 = 233$ .
Find $a_{5}$ : Finally, we find $a_{5}$ using the value of $a_{4}$ : $a_{5} = 5 \times a_{4} + 1 = 5 \times 233 + 1 = 1165 + 1 = 1166$ .