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

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 + \dots

(i i) 11 + 23 + 59 + 167 + \dots

Answered by math1967 last updated on 17/Jul/18

i) 4 + 6 + 9 + 13 + 18 + . . = S n

4 + 6 + 9 + 13 + \dots = S n

b y s u b t r a c t i n g

4 + {2 + 3 + 4 + 5 + \dots \dots (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} + \dots . 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 + \dots 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 + \dots (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 + \dots

11 + 23 + 59 + \dots .

11 + 12 (1 + 3 + 3^{2} + \dots .) = t_{n}

t_{n} = 6 \times 3^{n - 1} + 5

S_{n} = 6 (1 + 3 + 3^{2} + \dots .) + 5 n

= 6 \times \frac{3^{n} - 1}{3 - 1} + 5 n = 3 (3^{n} - 1) + 5 n

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