Respuesta :
[tex]\bf \stackrel{4^2}{16}+\stackrel{5^2}{25}+\stackrel{6^2}{36}+\stackrel{7^2}{49}+...\qquad \qquad \qquad \sum\limits_{i=4}^{\infty}~i^2[/tex]
Answer:
Hence, the sum using summation notation, assuming the suggested pattern continues [tex]16+25+36+49+......+n^2+.....[/tex] is:
[tex]\sum_{n=4}^{\infty}n^2[/tex]
Step-by-step explanation:
We have to write the sum using summation notation, assuming the suggested pattern continues:
[tex]16+25+36+49+......+n^2+.....[/tex]
Clearly we may also write this pattern as:
[tex]4^2+5^2+6^2+.....+n^2+......[/tex]
So, in terms of the summand it is written as:
[tex]\sum_{n=4}^{\infty}n^2[/tex]
( We have started our summation from 4 since the term in the summation starts with 16 which is 4^2 and goes to infinity )