3+8+15+24+35+... find sum of 50^(th) −term


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Question Number 104821 by bobhans last updated on 24/Jul/20

$3 + 8 + 15 + 24 + 35 + . . .$ $f i n d s u m o f 50^{t h} - t e r m$
	Answered by bemath last updated on 24/Jul/20

	$S_{50} = 3 C_{1}^{50} + 5 C_{2}^{50} + 2 C_{3}^{50}$ $= 150 + 5 \times 25 \times 49 + 2 \times \frac{50 \times 49 \times 48}{3 \times 2 \times 1}$ $= 45475 ★$
	Commented by bobhans last updated on 24/Jul/20

	$t h a n k y o u$
	Answered by 1549442205PVT last updated on 24/Jul/20
	$Since 8−3=5,15−8=7,24−15=9,35−24=11 and 5,7,9,11...form an arithmetic progress, u_n =an^2 +bn+c,so { ((u_1 =3=a+b+c(1))),((u_2 =8=4a+2b+c(2)⇒ { ((3a+b=5(4))),((5a+b=7(5))) :})),((u_3 =15=9a+3b+c(3))) :} ⇒2a=2⇒a=1⇒b=2,c=0 u_n =n^2 +2n⇒u_(50) =50^2 +2×50=2600 S_n =Σ_(k=1) ^(n) (k^2 +2k)=Σ_(k=1) ^(n) k^2 +2Σ_(l=1) ^(n) k=((n(n+1)(2n+1))/6)+n(n+1) =((n(n+1)(2n+7))/6).Therefore, S_(50) =((50×51×107)/6)=25×17×107=45475$
	$Since 8 - 3 = 5, 15 - 8 = 7, 24 - 15 = 9, 35 - 24 = 11$ $and 5, 7, 9, 11 . . . form an arithmetic progress,$ $u_{n} = {an}^{2} + bn + c, so$ ${\begin{cases} u_{1} = 3 = a + b + c (1) \\ u_{2} = 8 = 4 a + 2 b + c (2) \Rightarrow {\begin{cases} 3 a + b = 5 (4) \\ 5 a + b = 7 (5) \end{cases} \\ u_{3} = 15 = 9 a + 3 b + c (3) \end{cases}$ $\Rightarrow 2 a = 2 \Rightarrow a = 1 \Rightarrow b = 2, c = 0$ $u_{n} = n^{2} + 2 n \Rightarrow u_{50} = 50^{2} + 2 \times 50 = 2600$ $S_{n} = \underset{k = 1}{\overset{n}{Σ}} (k^{2} + 2 k) = \underset{k = 1}{\overset{n}{Σ}} k^{2} + 2 \underset{l = 1}{\overset{n}{Σ}} k = \frac{n (n + 1) (2 n + 1)}{6} + n (n + 1)$ $= \frac{n (n + 1) (2 n + 7)}{6} . Therefore,$ $S_{50} = \frac{50 \times 51 \times 107}{6} = 25 \times 17 \times 107 = 45475$
	Commented by bobhans last updated on 24/Jul/20

	$s i r t h e q u e s t i o n w a n t t o f i n d \underset{n = 1}{\sum^{50}} u_{n}$ $o y e s . . t h a n k y o u$
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