Sum of N squares minus sum of N integers

Algebra Level 2

Solve for x x such that

3333 x = ( 1 + 2 2 + 3 2 + + 10 0 2 ) ( 1 + 2 + 3 + + 100 ) 3333x = (1+2^2 + 3^2 + \cdots + 100^2) - ( 1+2+3+\cdots + 100)


The answer is 100.

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2 solutions

To generalize, let us consider:

S ( n ) = 1 2 + 2 2 + 3 2 + + 2 n ( 1 + 2 + 3 + + n ) = n ( n + 1 ) ( 2 n + 1 ) 6 n ( n + 1 ) 2 = n ( n + 1 ) ( 2 n + 1 3 ) 6 = ( n 1 ) n ( n + 1 ) 3 \begin{aligned} S(n) & = 1^2 + 2^2 + 3^2 + \cdots + 2^n - (1 + 2 + 3 + \cdots + n) \\ & = \frac {n(n+1)(2n+1)}6 - \frac {n(n+1)}2 \\ & = \frac {n(n+1)(2n+1-3)}6 \\ & = \frac {(n-1)n(n+1)}3 \end{aligned}

Therefore S ( 100 ) = 99 × 100 × 101 3 = 333300 x = 100 S(100) = \dfrac {99\times 100 \times 101}3 = 333300 \implies x = \boxed{100} .

Srinivasa Gopal
Mar 7, 2020

Sum of squares of first N natural numbers is equal to (N (N+1) (2N+1))/6

Sum of squares of first 100 natural numbers is equal to (100 101 201)/6 = 50 101 67 = 338350333300

Sum of first N natural numbers is equal to N (N+1)/2 = 100 101/2 = 5050

So (1+2^2 + 3^2 + 4^2 + 5^2+ 6^2......100^2) - ( 1+2+3+4+5+6......100) = 338350 - 5050 = 333300

3333 *x = 333300 solving for x yields x = 333300/3333 = 100.

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