The n^{th} term of the following series (1 ti | vedclass

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

Question

(C) The given series is $(1 	imes 3) + (3 	imes 5) + (5 	imes 7) + (7 	imes 9) + \dots$Observe the first factors of each term: $1, 3, 5, 7, \dots$

Vedclass · Accepted Answer

$(2n - 1)(2n + 1)$

Vedclass · Answer

(C) The given series is $(1 	imes 3) + (3 	imes 5) + (5 	imes 7) + (7 	imes 9) + \dots$Observe the first factors of each term: $1, 3, 5, 7, \dots$This is an arithmetic progression with first term $a = 1$ and common difference $d = 2$.The $n^{th}$ term of this sequence is $a_n = a + (n - 1)d = 1 + (n - 1)2 = 2n - 1$.Observe the second factors of each term: $3, 5, 7, 9, \dots$This is an arithmetic progression with first term $a = 3$ and common difference $d = 2$.The $n^{th}$ term of this sequence is $b_n = 3 + (n - 1)2 = 2n + 1$.Therefore,the $n^{th}$ term of the series is the product of the $n^{th}$ terms of these two sequences:$T_n = (2n - 1)(2n + 1)$.

The $n^{th}$ term of the following series $(1 \times 3) + (3 \times 5) + (5 \times 7) + (7 \times 9) + \dots$ will be

Similar Questions

Find the sum to $n$ terms of the series $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \dots$

If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

What is the sum of the first $10$ terms of the series $0.7 + 0.77 + 0.777 + \dots$?

$\sum_{k=1}^5 \frac{1^3+2^3+\ldots+k^3}{1+3+5+\ldots+(2 k-1)}$ is equal to (in $.5$)

If $1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots \text{ to } n \text{ terms} = n(n+1) f(n) - 3n$,then $f(1) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module