Solve the inequality 16

Mathematics

1 answer:

GarryVolchara [31]3 years ago

8 0

Answer:

I thought I knew how to do this but ig I forgot ill be back in a little with the anwser.

Step-by-step explanation:

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The sum of the triangular numbers is $\frac{n(n+1)(n+2)}{6}$

According to the question, the series of triangular numbers is given as in the form of the number of dots constituting the equilateral triangles i.e.
1, 3, 6, 10, 15 . . . . .n

<h3>what is triangular numbers?</h3>

triangular number are the sequence and series of the numbers and each number represents and constitute in visualization of series of equilateral triangle.

given series in the figure
1, 3, 6, 10, 15, . . . . . . ., n
each number represents the number of dots containing in triangle
now the series can be given by
$S = \frac{n(n+1)}{2}$ for n = 1, 2, 3, 4,. . . . .n
now , according to the question the sum of the series can be given as
⇒ ∑S
⇒ ∑ $\frac{n^2+n}{2}$
⇒ $\frac{1}{2}[\sum n^2+\sum n]\\\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}] \\\frac{n(n+1)}{4}[\frac{(2n+1)}{3}+1] \\\frac{n(n+1)}{4}[\frac{(2n+1+3)}{3}] \\\frac{n(n+1)}{4}[\frac{(2n+4)}{3}] \\\frac{n(n+1)}{4}2[\frac{(n+2)}{3}]\\\frac{n(n+1)(n+2)}{6}$