Permutated Pooled Pascal (Power of 7)

Algebra Level 3

( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) 1 + 7 ( 1 + 2 + 3 + 4 + 5 + 6 ) 2 + 21 ( 1 + 2 + 3 + 4 + 5 ) 3 + 35 ( 1 + 2 + 3 + 4 ) 4 + 35 ( 1 + 2 + 3 ) 5 + 21 ( 1 + 2 ) 6 + 7 ( 1 ) 7 = ? { (1+2+3+4+5+6+7) }^{ 1 } \\ + 7 { (1+2+3+4+5+6) }^{ 2 } \\ + 21{ (1+2+3+4+5) }^{ 3 } \\ + 35{ (1+2+3+4) }^{ 4 } \\ +35{ (1+2+3) }^{ 5 } \\ +21{ (1+2) }^{ 6 } \\ +7 { (1) }^{ 7 } \\ = \ ?


The answer is 711466.

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1 solution

( 1 + 2 + 3 + 4 + 5 + 6 + 7 ) 1 (1 + 2 + 3 + 4 + 5 + 6 + 7)^1

+ + 7 ( 1 + 2 + 3 + 4 + 5 + 6 ) 2 7(1 + 2 + 3 + 4 + 5 + 6)^2

+ + 21 ( 1 + 2 + 3 + 4 + 5 ) 3 21(1 + 2 + 3 + 4 + 5)^3

+ + 35 ( 1 + 2 + 3 + 4 ) 4 35(1 + 2 + 3 + 4)^4

+ + 35 ( 1 + 2 + 3 ) 5 35(1 + 2 + 3)^5

+ + 21 ( 1 + 2 ) 6 21(1 + 2)^6

+ + 7 ( 1 ) 7 7(1)^7

The above line of operations change to:

( 28 ) 1 + 7 ( 21 ) 2 + 21 ( 15 ) 3 + 35 ( 10 ) 4 + 35 ( 6 ) 5 + 21 ( 3 ) 6 + 7 ( 1 ) 7 (28)^1 + 7(21)^2 + 21(15)^3 + 35(10)^4 + 35(6)^5 + 21(3)^6 + 7(1)^7

And once again changes to:

28 + 3087 + 70875 + 350000 + 272160 + 15309 + 7 28 + 3087 + 70875 + 350000 + 272160 + 15309 + 7

And then, the final result is 711466 \boxed {711466}

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