Tk-k2-2k- then find -k-1nTk-

∑_k=1^nT_k=∑_k=1^nk^2+∑_k=1^n2^k=n ( n+1 ) ( 2n+1 )/6+2 ( 2^n 1 )/2 1=n ( n+1 ) ( 2n+1 )/6+2^n+1 1. ...

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Integration Using Substitution

Tk=k2+2k, the...

Question

$T_{k} = k^{2} + 2^{k},$ then find $\sum_{k = 1}^{n} T_{k} .$

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Solution

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$X = x$	$0$	1	2	3	4	5	6	7
$P (X = x)$	$0$	$k$	$2 k$	$2 k$	$3 k$	$K^{2}$	$2 k^{2}$	$7 k^{2} + k$

P (0 < X < 5) =

If $s e c θ + t a n θ = k, c o s θ =$

k

(k^{2} + 2 k)

f (n) = n \sum k = 1 {cosec}^{- 1} \sqrt{(k^{2} + 1) (k^{2} + 2 k + 2)}

lim n \to \infty \frac{12 f (n)}{π}

f (n) = n \sum k = 1 {cosec}^{- 1} \sqrt{(k^{2} + 1) (k^{2} + 2 k + 2)}

lim n \to \infty \frac{12 f (n)}{π}

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