∑i=knf(i)=∑i=k+pn+pf(i−p) We change i from the initial k to (k+p) \displaystyle \sum_{i=k}^{n}f(i) = \sum_{i=k+p}^{n+p}f(i-p)~~~~We~change~~ i~~ from~the~initial~~k~~to~~(k+p)~~ i=k∑nf(i)=i=k+p∑n+pf(i−p) We change i from the initial k to (k+p) and final from n to (n+p) while in the function (i−p) replaces i.\displaystyle and~~~final~~from~~n~~to~~(n+p)~~while~ in~the~function~~(i-p)~~replaces~~i.and final from n to (n+p) while in the function (i−p) replaces i. p is any +tive or −tive integer.\displaystyle ~~~~~~ p~ is~any~ +tive~ or~- tive~ integer. p is any +tive or −tive integer. Summation can be split as follows:−∑i=knf(i)=∑i=krf(i)+∑i=r+1nf(i)\displaystyle Summation~can~be~split~as~follows:- \sum_{i=k}^{n}f(i) = \sum_{i=k}^{r}f(i)+ \sum_{i=r+1}^{n}f(i) Summation can be split as follows:−i=k∑nf(i)=i=k∑rf(i)+i=r+1∑nf(i)
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Note by Niranjan Khanderia 7 years, 2 months ago
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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2^{34}
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\sum_{i=1}^3
\sin \theta
\boxed{123}
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