First slide

Sequence and Series

Question

If $1 + 5 + 12 + 22 + 35 + \dots to n terms = \frac{n^{2} (n + 1)}{2}$ n^th term of series is

Easy

A

$\frac{n (4 n - 1)}{3}$
B

$\frac{n (3 n - 1)}{2}$
C

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

$\frac{n (4 n + 1)}{3}$

Solution

Let
$\begin{array}{r} P (n) : 1 + 5 + 12 + 22 + 35 + \dots + (n terms) \\ = \frac{n^{2} (n + 1)}{2} \end{array}$
nth term of LHS = P(n) - P(n - 1)
$\begin{array}{l} \Rightarrow P (n) - P (n - 1) = \frac{n^{2} (n + 1)}{2} - \frac{(n - 1)^{2} n}{2} \\ \Rightarrow P (n) - P (n - 1) = \frac{n}{2} \{n^{2} + n - n^{2} + 2 n - 1\} \\ ∴ T_{n} = P (n) - P (n - 1) \\ = \frac{n}{2} (3 n - 1) \end{array}$

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