Find the sum:

$$\sum _{k=3}^{6}({k}^{2}-7)$$

$$\sum _{k=3}^{6}({k}^{2}-7)$$

Antwan Perez
2022-10-20
Answered

Find the sum:

$$\sum _{k=3}^{6}({k}^{2}-7)$$

$$\sum _{k=3}^{6}({k}^{2}-7)$$

You can still ask an expert for help

Hilfeform5c

Answered 2022-10-21
Author has **14** answers

Solved:

$$\sum _{k=3}^{6}({k}^{2}-7)\phantom{\rule{0ex}{0ex}}\Rightarrow (3{)}^{2}-7+(4{)}^{2}-7+(5{)}^{2}-7+(6{)}^{2}-7\phantom{\rule{0ex}{0ex}}\Rightarrow 2+9+18+29\phantom{\rule{0ex}{0ex}}\Rightarrow 58$$

$$\sum _{k=3}^{6}({k}^{2}-7)\phantom{\rule{0ex}{0ex}}\Rightarrow (3{)}^{2}-7+(4{)}^{2}-7+(5{)}^{2}-7+(6{)}^{2}-7\phantom{\rule{0ex}{0ex}}\Rightarrow 2+9+18+29\phantom{\rule{0ex}{0ex}}\Rightarrow 58$$

asked 2022-06-20

Real number in terms of infinite series

$A={a}_{0}+{\left(\frac{1}{{a}_{1}}\right)}^{k}+{\left(\frac{1}{{a}_{2}}\right)}^{k}+{\left(\frac{1}{{a}_{3}}\right)}^{k}+\dots $

Where $i\ge 1$ and the recurrence relation ${a}_{i+1}\ge {a}_{i}\ge 2$

$A={a}_{0}+{\left(\frac{1}{{a}_{1}}\right)}^{k}+{\left(\frac{1}{{a}_{2}}\right)}^{k}+{\left(\frac{1}{{a}_{3}}\right)}^{k}+\dots $

Where $i\ge 1$ and the recurrence relation ${a}_{i+1}\ge {a}_{i}\ge 2$

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$\underset{n\to +\mathrm{\infty}}{lim}\sum _{k=1}^{n}\frac{1}{k(k+1)\cdots (k+m)}\phantom{\rule{2em}{0ex}}(m=1,2,3,\cdots )$

$\underset{n\to +\mathrm{\infty}}{lim}\sum _{k=1}^{n}\frac{1}{k(k+1)\cdots (k+m)}\phantom{\rule{2em}{0ex}}(m=1,2,3,\cdots )$

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