find the nth term and the sum to n termof the following seried (i) 4+6+9+13+18+... (ii) 11+23+59+167+...


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Question Number 40218 by scientist last updated on 17/Jul/18

$f i n d t h e n t h t e r m a n d t h e s u m t o n t e r m o f t h e f o l l o w i n g s e r i e d$ $(i) 4 + 6 + 9 + 13 + 18 + . . .$ $(i i) 11 + 23 + 59 + 167 + . . .$
	Answered by math1967 last updated on 17/Jul/18
	$i)4+6+9+13+18+..=Sn 4+6+9+13+...=Sn by subtracting 4+{2+3+4+5+......(n−1)term}=t_n ∴t_n =4+(((n−1)(n+2))/2)=((n^2 +n+6)/2) now S_n =(1/2)(1^2 +2^2 +....n^2 )+(1/2)(1+2+..)+3n =(1/(12))(n+1)(2n+1)+(1/4)n(n+1)+3n$
	$i) 4 + 6 + 9 + 13 + 18 + . . = S n$ $4 + 6 + 9 + 13 + . . . = S n$ $b y s u b t r a c t i n g$ $4 + {2 + 3 + 4 + 5 + . . . . . . (n - 1) t e r m} = t_{n}$ $∴ t_{n} = 4 + \frac{(n - 1) (n + 2)}{2} = \frac{n^{2} + n + 6}{2}$ $n o w S_{n} = \frac{1}{2} (1^{2} + 2^{2} + . . . . n^{2}) + \frac{1}{2} (1 + 2 + . .) + 3 n$ $= \frac{1}{12} (n + 1) (2 n + 1) + \frac{1}{4} n (n + 1) + 3 n$
	Commented by scientist last updated on 17/Jul/18

	$f r o m t h i s s o l u t i o n i g u e s s, s u p p o s e (1 + 2 + 3 + 4 + . . . n)$ $= \frac{n (n + 1)}{2}$ $s o w h e r e d i d u n o w g e t (n - 1) (n + 2) i n s t e a d (n - 1) (n)$
	Commented by math1967 last updated on 18/Jul/18
	$4+{2+3+4+...(n−1)term} 4+(((n−1){2×2+(n−1−1)×1})/2) [using S_(n ) =(n/2){2a+(n−1)d}] 4+(((n−1)(4+n−2))/2) 4+(((n−1)(n+2))/2)=((8+n^2 +n−2)/2)=((n^2 +n+6)/2)$
	$4 + {2 + 3 + 4 + . . . (n - 1) t e r m}$ $4 + \frac{(n - 1) {2 \times 2 + (n - 1 - 1) \times 1}}{2}$ $[u s i n g S_{n} = \frac{n}{2} {2 a + (n - 1) d}]$ $4 + \frac{(n - 1) (4 + n - 2)}{2}$ $4 + \frac{(n - 1) (n + 2)}{2} = \frac{8 + n^{2} + n - 2}{2} = \frac{n^{2} + n + 6}{2}$
	Commented by scientist last updated on 18/Jul/18

	$G o t i t s i r, s h u k r a h$
	Answered by math1967 last updated on 17/Jul/18

	$i i) 11 + 23 + 59 + 167 + . . .$ $11 + 23 + 59 + . . . .$ $11 + 12 (1 + 3 + 3^{2} + . . . .) = t_{n}$ $t_{n} = 6 \times 3^{n - 1} + 5$ $S_{n} = 6 (1 + 3 + 3^{2} + . . . .) + 5 n$ $= 6 \times \frac{3^{n} - 1}{3 - 1} + 5 n = 3 (3^{n} - 1) + 5 n$
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