What is this equal to?

Mathematics

1 answer:

Sauron [17]3 years ago

8 0

ANSWER

$\sum_{n=1} ^{32} (4n + 1) = 2144$

EXPLANATION

The given series is

$\sum_{n=1} ^{32} (4n + 1)$

The first term in this series is

$a_1=4(1) + 1 = 5$

The last term is

$l = 4(32) + 1 = 129$

The sum of the first n terms is

$S_n= \frac{n}{2} (a + l)$

The sum of the first 32 terms is

$S_ {32} = \frac{32}{2} (5 + 129)$

$S_ {32} =16 \times 134$

$S_ {32} =2144$

Therefore,

$\sum_{n=1} ^{32} (4n + 1) = 2144$

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Answer:

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Step-by-step explanation:

Given definite integral is:

$\int\limits^7_4 {\frac{x}{2}+x^{3}} \, dx \\f(x)=\frac{x}{2}+x^{3}---(1)\\\Delta x=\frac{b-a}{n}\\\\\Delta x=\frac{7-4}{n}=\frac{3}{n}\\\\x_{i}=a+\Delta xi\\a= Lower Limit=4\\\implies x_{i}=4+\frac{3}{n}i---(2)\\\\then\\f(x_{i})=\frac{x_{i}}{2}+x_{i}^{3}$

Substituting (2) in above

$f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$