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Algebra Level 2

The $n^{th}$ term of the series $1^2 , ( 1^2 + 2^2 ) , ( 1^2 + 2^2 + 3^2 )$ , $\ldots$ is?

\frac{ n(n + 1)(2n + 1) }{6}

\frac{ n(n + 1) }{2}

\frac{n^2 (n + 1)^2 }{2}

n^2 + (n - 1)^2 + (n - 2)^2

1 solution

Ruslan Abdulgani
Mar 11, 2015

Un = an3 + bn2 + cn + d, a+b+c+d = 1
8a+4b+2c+d=5 7a+3b+c=4 12a+2b=5 27a+9b+3c+d=14 19a+5b+c=9 18a+2b=7, 64a+16b+4c+d=30 37a+7b+c=16,
a=1/3, b=1/2, c=1/6, d=0 Un = 1/6n(2n2 + 3n +1) = 1/6(2n+1)(n+1)

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