The sum ∑_(r=2)^∞▒1/(r²-1) is equal to:a)1 b)3/4 c)4/3 d)None of these

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Find the value of the expression ∑_(i=1)^n▒∑_(j=1)^i▒∑_(k=1)^j▒1
Sequence and series class 11 JEE
The different numbers occurring in any particular sequence are known as its terms. The terms of a sequence are denoted by

a1, a2, a3,….,an

If a sequence has a finite number of terms then it’s known as a finite sequence. A sequence is termed as infinite if it is not having a definite number of terms. The nth term of an AP is given by

a + (n-1) d.

Between any two numbers ‘a’ and ‘b’, n numbers can be inserted such that the resulting sequence is an Arithmetic Progression. A1 , A2 , A3,……,An be n numbers between a and b such that a, A1 , A2 , A3,……,An, b is in A.P.

Here, a is the 1st term and b is (n+2)th term. Therefore,

b = a + d[(n + 2) – 1] = a + d (n + 1).

Hence, common difference (d) = (b-a)/(n+1)

Now, A1= a+d= a+((b-a)/(n+1)),

A2= a+2d = a+ ((2(b-a)/(n+1)),

An =a+nd= a+ ((n(b-a)/(n+1))}

The nth term of a geometric progression is given by an = arn-1
Sequence and Series Formulas
List of some basic formula of arithmetic progression and geometric progression are,

Arithmetic Progression Geometric Progression
Sequence a, a+d, a+2d,……,a+(n-1)d,…. a, ar, ar2,….,ar(n-1),…
Common Ratio Successive term – Preceding term
Common difference = d = a2 – a1

Successive term/Preceding term
Common ratio = r = ar(n-1)/ar(n-2)

General Term (nth Term) an = a + (n-1)d an = ar(n-1)
nth term from the last term an = l – (n-1)d an = 1/r(n-1)
Sum of first n terms sn = n/2(2a + (n-1)d) sn = a(rn)/(1-r) if r 1
sn = a(rn -1)/(r – 1)………(2) if r greater than 1
Sequence and Series Examples
Question 1: If 4,7,10,13,16,19,22……is a sequence, Find:

Common difference
nth term
21st term
Solution: Given sequence is, 4,7,10,13,16,19,22……

a) The common difference = 7 – 4 = 3

b) The nth term of the arithmetic sequence is denoted by the term Tn and is given by Tn = a + (n-1)d, where “a” is the first term and d, is the
common difference.
Tn = 4 + (n – 1)3 = 4 + 3n – 3 = 3n + 1
c) 21st term as: T21 = 4 + (21-1)3 = 4+60 = 64.

Question 2: Consider the sequence 1,4,16,64,256,1024….. Find the common ratio and 9th term.

Solution: The common ratio (r) = 4/1 = 4

The preceding term is multiplied by 4 to obtain the next term.

The nth term of the geometric sequence is denoted by the term Tn and is given by Tn = ar(n-1)
where a is the first term and r is the common ratio.

Here a = 1, r = 4 and n = 9

So, 9th term is can be calculated as T9 = 1* (4)(9-1)= 48 = 65536.
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