If displaystyle∑ i =0∞ displaystyle∑ j =0∞ (1/ a i ⋅ a j )=(λ a 2/( a -1)2( a +1)) where i ≠ j and a 1 then possible values of λ may be

Question

If displaystyle∑ i =0∞ displaystyle∑ j =0∞ (1/ a i ⋅ a j )=(λ a 2/( a -1)2( a +1)) where i ≠ j and a >1 then possible values of λ may be (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

When no restriction on i and j S = displaystyle∑ i =0∞ displaystyle∑ j =0∞ (1/ a i ⋅ a j )=(1+(1/ a )+(1/ a 2)+ ldots ∞)2=( a 2/( a -1)2) i ≠ j ⇒ S =(λ a 2/( a -1)2( a +1)) ⇒ λ>2

Q. If $i = 0 \sum \infty j = 0 \sum \infty \frac{1}{a ^{i} \cdot a ^{j}} = \frac{λ a ^{2}}{( a - 1 ) ^{2} ( a + 1 )}$ where $i \neq = j$ and $a > 1$ then possible values of $λ$ may be

1

2

3

4

When no restriction on i and j
$S = i = 0 \sum \infty j = 0 \sum \infty \frac{1}{a ^{i} \cdot a ^{j}} = (1 + \frac{1}{a} + \frac{1}{a ^{2}} + \dots \infty)^{2} = \frac{a ^{2}}{( a - 1 ) ^{2}} i \neq = j$
$\Rightarrow S = \frac{λ a ^{2}}{( a - 1 ) ^{2} ( a + 1 )} \Rightarrow λ > 2$

Q. If i=0∑∞​j=0∑∞​ai⋅aj1​=(a−1)2(a+1)λa2​ where i=j and a>1 then possible values of λ may be

1

2

3

4

When no restriction on i and j S=i=0∑∞​j=0∑∞​ai⋅aj1​=(1+a1​+a21​+…∞)2=(a−1)2a2​i=j ⇒S=(a−1)2(a+1)λa2​⇒λ>2

Q. If $i = 0 \sum \infty j = 0 \sum \infty \frac{1}{a ^{i} \cdot a ^{j}} = \frac{λ a ^{2}}{( a - 1 ) ^{2} ( a + 1 )}$ where $i \neq = j$ and $a > 1$ then possible values of $λ$ may be

When no restriction on i and j
$S = i = 0 \sum \infty j = 0 \sum \infty \frac{1}{a ^{i} \cdot a ^{j}} = (1 + \frac{1}{a} + \frac{1}{a ^{2}} + \dots \infty)^{2} = \frac{a ^{2}}{( a - 1 ) ^{2}} i \neq = j$
$\Rightarrow S = \frac{λ a ^{2}}{( a - 1 ) ^{2} ( a + 1 )} \Rightarrow λ > 2$