Find the sum of the series $1^2+3^2+5^2+\cdots+(2n-1)^2$
It is a mathematical question and I can't get it answer. Can anyone help me with this question?
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2Do you know the formula for a sum of consecutive squares? Sum all squares up to $(2n)^2$ and subtract teh even ones -- namely $2^2$ times the sum of all squares up to $n^2$. – hmakholm left over Monica Apr 11 '17 at 10:31
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1This has to be a duplicate. – Apr 11 '17 at 10:34
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1Write out the first few values. Guess the general form. Prove it by induction. – anomaly Apr 11 '17 at 10:34
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Henning's comment gives: $$\begin{align}1^2+2^2+\cdots+(2n)^2 &= [1^2+3^2+\cdots+(2n-1)^2] + [2^2+4^2+\cdots+(2n)^2] \ & = [1^2+3^2+\cdots+(2n-1)^2] + 2^2 [1^2+2^2+\cdots+n^2]\end{align}$$ – Frenzy Li Apr 11 '17 at 10:39
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@A---B Here are a few duplicate targets (there must be more). Pick your favorite, ordered chronologically, newest first: 2230785, 533827, 437835, and 7-year-old 72636. – Lee David Chung Lin Jan 11 '19 at 07:29
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Possible duplicate of Compute $1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2$ by mathematical induction – Lee David Chung Lin Jan 11 '19 at 07:30
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Hint:
Notice We can write your sum as
$$ S = \sum_{i=1}^n (2i-1)^2 = \sum_i (4i^2 - 4i + 1) = 4 \sum_i i^2 - 4 \sum_i i + n$$
ILoveMath
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These formulas may help you: $$\sum\limits_{k=1}^{n} k^2=\frac {n (n+1)(2n+1)}{6} $$ $$\sum\limits_{k=1}^{n} k=\frac {n (n+1)}{2} $$ $$\sum\limits_{k=1}^{n} c=cn$$ Your result should then be something like $$\frac {(2n-1)2n (2n+1)}{6} $$
mxian
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Using finite calculus we can sum polynomials using what's called falling powers:
\begin{align} \sum_{x=0}^{n-1} (2x+1)^2 &= \sum_{x=0}^{n-1} (4x^2+4x+1) = \sum_{x=0}^{n-1} [4x(x-1) + 8x + 1] \\ &= \sum_{x=0}^{n-1} [4x^{\underline2} + 8x^{\underline1} + 1] = \left.\frac{4}{3}x^{\underline3}+4x^{\underline2}+x^{\underline1}\right|_{0}^{n} \\ &= \left.\frac{4}{3}x(x-1)(x-2)+4x(x-1)+x\right|_{0}^{n} \\ &= \frac{4}{3}n(n-1)(n-2)+4n(n-1)+n \\ &= \frac{n}{3}(4n^2-1). \end{align}
Frenzy Li
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