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S-1-k-1-n-16n-16k-16n-16k-S-2-k-1-n-16k-16-16k-16-lim-n-S-1-S-2-n-2-




Question Number 61510 by Tony Lin last updated on 03/Jun/19
S_1 =Σ_(k=1) ^n (√((16n−16k)(16n+16k)))  S_2 =Σ_(k=1) ^n (√((16k−16)(16k+16)))  lim_(n→∞) ((S_1 +S_2 )/n^2 )=?
$${S}_{\mathrm{1}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{\left(\mathrm{16}{n}−\mathrm{16}{k}\right)\left(\mathrm{16}{n}+\mathrm{16}{k}\right)} \\ $$$${S}_{\mathrm{2}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{\left(\mathrm{16}{k}−\mathrm{16}\right)\left(\mathrm{16}{k}+\mathrm{16}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{S}_{\mathrm{1}} +{S}_{\mathrm{2}} }{{n}^{\mathrm{2}} }=? \\ $$

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