sigma

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sigma

Mathematics
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\[\sum_{x=1}^{25}\] K^2 -3k +4 what is the formula to solve if u dont have calculator
please help
\[\sum_{x=1}^{25}k^2-3\sum_{x=1}^{25}k+4\sum_{x=1}^{25}1\]

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Other answers:

imran please go ahead, i have to get back to work
but this one is pretty easy \[4\sum_{k=1}^{25}1=4\times 25=100\]
sum 1 to n of (1) is equal to n sum 1 to n of (i) is equal to n(n+1)/2 sum 1 to n of (i^2) is equal to n(n+1)(2n+1)/6 um i think.
yes, agreed^
@curry you got this?
\[\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\] so \[-3\sum_{k=1}^{25}k=-3\times \frac{25\times 26}{2}\]
kind a
\[\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}\] so \[\sum_{k=1}^{25}k^2=\frac{25(26)(51)}{6}\]
now you have all the numbers you need, grind it out with a calculator
should get 4650 if you do it right http://www.wolframalpha.com/input/?i=sum+k+%3D+1+to+25+%28k^2-3k%2B4%29

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