The n^{th} term of the series frac{2}{1!} + f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given series is $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$Observe the numerators: $2, 7, 15, 26, \dots$Let $a_n$ be

Vedclass · Accepted Answer

$\frac{n(3n + 1)}{2(n!)}$

Vedclass · Answer

(B) The given series is $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$Observe the numerators: $2, 7, 15, 26, \dots$Let $a_n$ be the numerator of the $n^{th}$ term. The differences between consecutive terms are $5, 8, 11, \dots$,which form an Arithmetic Progression $(AP)$ with first term $a = 5$ and common difference $d = 3$.The $n^{th}$ term of the numerator sequence can be expressed as the sum of the first $n$ terms of this $AP$ plus the initial value:$a_n = 2 + \sum_{k=0}^{n-1} (5 + 3k) = 2 + \frac{n}{2} [2(5) + (n-1)3] = 2 + \frac{n}{2} [10 + 3n - 3] = 2 + \frac{n(3n + 7)}{2} = \frac{4 + 3n^2 + 7n}{2}$.Alternatively,looking at the pattern provided in the options:$T_n = \frac{2 + 5 + 8 + \dots + (3n - 1)}{n!} = \frac{\frac{n}{2}[2(2) + (n-1)3]}{n!} = \frac{\frac{n}{2}[4 + 3n - 3]}{n!} = \frac{n(3n + 1)}{2(n!)}$.

The $n^{th}$ term of the series $\frac{2}{1!} + \frac{7}{2!} + \frac{15}{3!} + \frac{26}{4!} + \dots$ is

Similar Questions

If the sides of a right-angled triangle are in $A.P.$,then the sides are proportional to

Let $a_1, a_2, a_3, \dots$ be an $A.P.$ such that $\frac{a_1 + a_2 + \dots + a_p}{a_1 + a_2 + \dots + a_q} = \frac{p^3}{q^3}$,where $p \neq q$. Then $\frac{a_6}{a_{21}}$ is equal to:

The four arithmetic means between $3$ and $23$ are

Consider an infinite $G.P.$ with first term $a$ and common ratio $r$. If its sum is $4$ and the second term is $3/4$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module