Identically equal

There is one non-negative integers x x that satisfy

x 3 + ( x + 1 ) 3 + ( x + 2 ) 3 + + ( x + n ) 3 = ( x + ( x + 1 ) + ( x + 2 ) + + ( x + n ) ) 2 \small x^{3}+(x+1)^{3}+(x+2)^{3}+\cdots +(x+n)^{3}=(x+(x+1)+(x+2)+\cdots +(x+n))^{2}

For any positive integer n > 1 n>1

Find that x x and write the value of x x as answer.


The answer is 0.

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1 solution

Agent T
May 25, 2021

Summation of n terms = n ( n + 1 ) 2 \dfrac{n(n+1)}{2}

And summation of first n 3 n^{3} terms= ( n ( n + 1 ) 2 ) 2 \left(\dfrac{n(n+1)}{2}\right) ^{2}

This will only be valid here when x = 0 \boxed{\large{x=0}}

Boom!

Btw what is the relevance of the title? Just curious...

Agent T - 2 weeks, 4 days ago

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Nothing can't think of any title

Monu Kumar - 2 weeks, 4 days ago

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I've a suggestion: identically equal \boxed{\text{identically equal}}

Agent T - 2 weeks, 4 days ago

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@Agent T I edited. Ok thank you

Monu Kumar - 2 weeks, 4 days ago

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@Monu Kumar Haha I'm honored !

Agent T - 2 weeks, 4 days ago

Man, what happened to your ultra-cool Sunglasses Avatar, Agent-Tee???!!!

tom engelsman - 2 weeks, 3 days ago

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Haha! Lemme change back to my ultra cool sunglasses avatar ;)

Agent T - 2 weeks, 3 days ago

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@Agent T Keep those weird physics contraption probs comin' too!!! :) Have a good day....

tom engelsman - 2 weeks, 3 days ago

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@Tom Engelsman Hey I have posted one more , hope you like it :D ! You too have an amazingly awesomee day!!

Agent T - 2 weeks, 3 days ago

Wait... x=1 works too!

William Ly - 2 weeks, 3 days ago

And x=1 is incorrect for some reason...

William Ly - 2 weeks, 3 days ago

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