Answer:

(41/ 8) e – (19 / 8) e-1 – 10

IF ANY MISTAKE IN EXPLAINATION REALLY SORRY

Explaination:

Put 2n + 1 = r, where r = 3, 5, 7

therefore n^ 2 +6n+10 (2 n+1)! = ( r-1 2 )^ 2 +3r-3+10 r! = r^ 2 +10r+2 9 4(r!)

Now Sigma r=3,5,7*** r(r-1)+11r+29 4(r!) = 1 4 Sigma r=3,5,7...( 1 (r-2)! + 11 (r-1)! + 29 r! )

= 1 4 \ ( 1 1! + 1 3! + 1 5! +....)+11( 1 2! + 1 4! + 1 6! +...)+29( 1 3! + 1 5! + 1 7! +.....)\

= 1 4 \ e- 1 e z +11( e+ 1 e z z )+29( e- 1 e -z z )\

= 1 8 \ e- 1 e +11e+ 11 e -22+29e- 29 e -58\

= 1 8 \ 41e- 19 e -80\

n = [r – 1] / 2

Verified answer

Answer:

[tex]\boxed{ \sf{ \:\bf \: \displaystyle \sum^ \infty _{n=1} \bf \: \frac{ {n}^{2} + 6n + 10 }{(2n + 1)!} = \dfrac{1}{8}\bigg(41e - 19{e}^{ - 1} - 80 \bigg) \: }} \\ \\ [/tex]

Step-by-step explanation:

Given expression is

[tex]\sf \: \displaystyle \sum^ \infty _{n=1} \: \dfrac{ {n}^{2} + 6n + 10 }{(2n + 1)!} \\ \\ [/tex]

Let first evaluate the expression

[tex]\sf \: \dfrac{ {n}^{2} + 6n + 10 }{(2n + 1)!} \\ \\ [/tex]

[tex]\sf \: = \dfrac{ 4{n}^{2} + 24n + 40 }{4 \: (2n + 1)!} \\ \\ [/tex]

[tex]\sf \: = \dfrac{ 4{n}^{2} + 4n + 20n + 1 + 39}{4 \: (2n + 1)!} \\ \\ [/tex]

[tex]\sf \: = \dfrac{ 4{n}^{2} + 4n + 1 + 20n +10 + 29}{4 \: (2n + 1)!} \\ \\ [/tex]

[tex]\sf \: = \dfrac{ {(2n + 1)}^{2} + 10(2n +1)+ 29}{4 \: (2n + 1)!} \\ \\ [/tex]

[tex]\sf \: = \frac{1}{4} \bigg(\dfrac{{(2n + 1)}^{2}}{\: (2n + 1)!} + \dfrac{10(2n + 1)}{(2n + 1)!} + \dfrac{29}{(2n + 1)!} \bigg) \\ \\ [/tex]

[tex]\sf \: = \frac{1}{4} \bigg(\dfrac{{(2n + 1)}}{(2n)!} + \dfrac{10}{2n!} + \dfrac{29}{(2n + 1)!} \bigg) \\ \\ [/tex]

[tex]\sf \: = \frac{1}{4} \bigg(\dfrac{{2n}}{(2n)!} + \dfrac{1}{2n!} + \dfrac{10}{2n!} + \dfrac{29}{(2n + 1)!} \bigg) \\ \\ [/tex]

[tex]\sf \: = \frac{1}{4} \bigg(\dfrac{{1}}{(2n - 1)!} + \dfrac{11}{2n!} + \dfrac{29}{(2n + 1)!} \bigg) \\ \\ [/tex]

So, Given expression

[tex]\sf \: \displaystyle \sum^ \infty _{n=1} \: \frac{ {n}^{2} + 6n + 10 }{(2n + 1)!} \\ \\ [/tex]

can be rewritten as

[tex]\sf \: = \displaystyle \sum^ \infty _{n=1} \: \frac{1}{4} \bigg(\dfrac{{1}}{(2n - 1)!} + \dfrac{11}{2n!} + \dfrac{29}{(2n + 1)!} \bigg) \\ \\ [/tex]

[tex]\sf \: =\frac{1}{4} \displaystyle \sum^ \infty _{n=1} \: \dfrac{{1}}{(2n - 1)!} +\frac{11}{4}\displaystyle \sum^ \infty _{n=1} \dfrac{1}{2n!} + \frac{29}{4}\displaystyle \sum^ \infty _{n=1}\dfrac{1}{(2n + 1)!} \\ \\ [/tex]

[tex]\sf \: = \dfrac{1}{4}\bigg( \dfrac{e - {e}^{ - 1} }{2} \bigg) + \dfrac{11}{4} \bigg( \dfrac{e + {e}^{ -1} }{2} - 1 \bigg) + \dfrac{29}{4}\bigg( \dfrac{e - {e}^{ - 1} }{2} - 1\bigg) \\ \\ [/tex]

[tex]\sf \: = \dfrac{1}{4}\bigg( \dfrac{e - {e}^{ - 1} }{2} \bigg) + \dfrac{11}{4} \bigg( \dfrac{e + {e}^{ -1} - 2 }{2} \bigg) + \dfrac{29}{4}\bigg( \dfrac{e - {e}^{ - 1} - 2}{2} \bigg) \\ \\ [/tex]

[tex]\sf \: = \dfrac{1}{8}\bigg(e - {e}^{ - 1} + 11e + 11{e}^{ - 1} - 22 + 29e - 29{e}^{ - 1} - 58 \bigg) \\ \\ [/tex]

[tex]\sf \: = \dfrac{1}{8}\bigg(41e - 19{e}^{ - 1} - 80 \bigg) \\ \\ [/tex]

Hence,

[tex]\boxed{ \sf{ \:\bf \: \displaystyle \sum^ \infty _{n=1} \bf \: \frac{ {n}^{2} + 6n + 10 }{(2n + 1)!} = \dfrac{1}{8}\bigg(41e - 19{e}^{ - 1} - 80 \bigg) \: }} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Formulae Used

[tex]\sf \: e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + - - - \\ \\ [/tex]

[tex]\sf \: {e}^{ - 1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} + - - - \\ \\ [/tex]

[tex]\sf \: e + {e}^{ - 1} = 2\bigg(1 + \frac{1}{2!} + \frac{1}{4!} + - - - \bigg) \\ \\ [/tex]

[tex]\sf \: e - {e}^{ - 1} = 2\bigg(1 + \frac{1}{3!} + \frac{1}{5!} + - - - \bigg) \\ \\ [/tex]

[tex]\sf \: \displaystyle \sum^ \infty _{n=1}\frac{1}{2n!} = \frac{1}{2!} + \frac{1}{4!} + - - - = \dfrac{e + {e}^{ - 1} }{2} - 1 \\ \\ [/tex]

[tex]\sf \: \displaystyle \sum^ \infty _{n=1}\frac{1}{(2n - 1)!} =1 + \frac{1}{3!} + \frac{1}{5!} + - - - = \dfrac{e - {e}^{ - 1} }{2} \\ \\ [/tex]

[tex]\sf \: \displaystyle \sum^ \infty _{n=1}\frac{1}{(2n + 1)!} = \frac{1}{3!} + \frac{1}{5!} + - - - = \dfrac{e - {e}^{ - 1} }{2} - 1\\ \\ [/tex]