m = 1 ∑ ∞ n = 1 ∑ ∞ 3 m ( n ⋅ 3 m + m ⋅ 3 n ) m 2 n
Simplify the expression above.
If your answer comes in the form of b a , where a and b are coprime positive integers, then enter a + b as your answer.
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why this method works?
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Because since m and n both cover the same interval, they're interchangeable in the sum.
Replace 'm' with 'n' and 'n' with 'm' ,
Now add these two expressions,
Then we get terms which will be in A.G.P.,
So, by simplifying we get answer to be 9 / 3 2 .
So a + b = 4 1 .
It would be 9 / 3 2 . Please correct it.
Same approach, did it above :)
It would be nice if you provide an explanation as to why the method you mentioned works :)
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S = m = 1 ∑ ∞ n = 1 ∑ ∞ 3 m ( n ⋅ 3 m + m ⋅ 3 n ) m 2 n
S = 2 1 ⎣ ⎡ m = 1 ∑ ∞ n = 1 ∑ ∞ 3 m ( n ⋅ 3 m + m ⋅ 3 n ) m 2 n + 3 n ( m ⋅ 3 n + n ⋅ 3 m ) m n 2 ⎦ ⎤
S = 2 1 ⎣ ⎡ m = 1 ∑ ∞ n = 1 ∑ ∞ 3 m + n ( n ⋅ 3 m + m ⋅ 3 n ) 3 n ⋅ m 2 n + 3 m ⋅ m n 2 ⎦ ⎤
S = 2 1 ⎣ ⎡ m = 1 ∑ ∞ n = 1 ∑ ∞ 3 m + n ( n ⋅ 3 m + m ⋅ 3 n ) m n ( n ⋅ 3 m + m ⋅ 3 n ) ⎦ ⎤
S = 2 1 m = 1 ∑ ∞ n = 1 ∑ ∞ 3 m + n m n
S = 2 1 m = 1 ∑ ∞ 3 m m n = 1 ∑ ∞ 3 n n
Note that ∑ n = 0 ∞ x n = 1 − x 1 , for ∣ x ∣ < 1 . Differentiating w.r.t x and multiplying by x on both sides, one gets that ∑ n = 0 ∞ n x n = ( 1 − x ) 2 x .
So:
S = 2 1 [ ( 1 − 1 / 3 ) 2 1 / 3 ⋅ ( 1 − 1 / 3 ) 2 1 / 3 ]
S = 3 2 9
Then:
a = 9 , b = 3 2 , a + b = 4 1