a-1-0-a-n-27-a-n-1-n-1-k-1-m-a-k-

Question Number 2575 by prakash jain last updated on 22/Nov/15

a_{1} = 0

a_{n} = 27 \times a_{n - 1} + (n - 1)

\underset{k = 1}{\sum^{m}} a_{k} = ?

Commented by prakash jain last updated on 22/Nov/15

a_{1} = 0

a_{2} = 1

a_{3} = 27 + 2

a_{4} = 27^{2} + 2.27 + 3

a_{5} = 27^{3} + 2 . 27^{2} + 3 \cdot 27 + 4

Commented by Yozzi last updated on 22/Nov/15

G e n e r a t i n g f u n c t i o n i s a p o s s i b l e

a p p r o a c h t o f i n d i n g a_{n} I t h i n k .

Commented by prakash jain last updated on 22/Nov/15

This sum is based on Q 2545 . a_{n} being

the quotient when 3^{3 n} - 26 n - 1 is divided

by 676 .

a_{n} should evaluate to \frac{3^{3 n} - 26 n - 1}{676}

Commented by Rasheed Soomro last updated on 23/Nov/15

W h a t a d e e p o b s e r v e r y o u a r e!!!

Answered by RasheedAhmad last updated on 23/Nov/15

L e t S = \underset{k = 1}{\sum^{m}} a_{k}

S = s_{1} + s_{2} + s_{3} + \dots + s_{m}

1 + 2 + \dots + n = \frac{n (n + 1)}{2} [F o r m u l a f o r s_{i}]

where

s_{1} = 1 + 2 + \dots (m - 1) terms = \frac{m (m - 1)}{2}

s_{2} = 27 (1 + 2 + \dots . (m - 2) terms = \frac{(m - 2) (m - 1)}{2}

s_{3} = 27^{2} (1 + 2 + \dots (m - 3) t e r m s

= \frac{(m - 3) (m - 2)}{2}

\dots .

s_{m} = 27^{m - 1} (1 + 2 + \dots 0 t e r m s = 0

\dots

Continue

Commented by Filup last updated on 23/Nov/15

I believe if I have evaluated correctly :

S = \underset{k = 1}{\sum^{n}} (27^{k - 1} (\underset{h = 1}{\sum^{m - k}} h))

S = \underset{k = 1}{\sum^{n}} \underset{h = 1}{\sum^{m - k}} 27^{k - 1} h

S = \underset{k = 1}{\sum^{n}} (27^{k - 1} (\frac{1}{2} (m - k) (1 + m - k)))

s_{1} = 1 + 2 + \dots + (m - 1)

s_{2} = 27 (1 + 2 + \dots + (m - 2))

s_{3} = 27^{2} (1 + 2 + \dots + (m - 3))

⋮

s_{n} = 27^{n - 1} (1 + 2 + \dots + (m - n))

∴ \underset{i = 1}{\sum^{n}} s_{i} = \frac{1}{2} \underset{k = 1}{\sum^{n}} (27^{k - 1} (m - k) (m - k + 1))

c o n t i n u e

Answered by prakash jain last updated on 23/Nov/15

a_{n} = 27^{n - 2} + 2 \cdot 27^{n - 3} + \dots + (n - 2) \cdot 27 + (n - 1)

t_{n} = 27^{n - 2} + 2 \cdot 27^{n - 3} + 3 \cdot 27^{n - 2} + \dots + (n - 2) \cdot 27

\frac{t_{n}}{27} = 27^{n - 3} + 2 \cdot 27^{n - 2} + \dots + (n - 3) \cdot 27 + (n - 2)

t_{n} - \frac{t_{n}}{27} = 27^{n - 2} + 27^{n - 3} + \dots . + 27 - (n - 2)

RHS black color is GP

\frac{26 t_{n}}{27} = \frac{27 (27^{n - 2} - 1)}{26} - n + 2

\frac{26}{27} t_{n} = \frac{27^{n - 1} - 27 - 26 n + 2 \cdot 26}{26}

t_{n} = \frac{27^{n} - 27 \cdot 27 - 26 \cdot 27 n + 2 \cdot 26 \cdot 27}{26 \cdot 26}

a_{n} = \frac{27^{n} - 27 \cdot 27 - 26 \cdot 27 n + 2 \cdot 26 \cdot 27}{26 \cdot 26} + n - 1

a_{n} = \frac{3^{3 n} - 26 \cdot 27 n + 26 \cdot 26 n - 27 + 2 \cdot 26 \cdot 27 - 26 \cdot 26}{676}

a_{n} = \frac{3^{3 n} - 26 n - 27.27 + 26 \cdot 27 + 26.27 - 26 \cdot 26}{676}

a_{n} = \frac{3^{3 n} - 26 n - 27 (27 - 26) + 26 (27 - 26)}{676}

a_{n} = \frac{3^{3 n} - 26 n - 1}{676}

\underset{i = 1}{\sum^{m}} a_{i} = \frac{1}{676} [\underset{i = 1}{\sum^{m}} 3^{3 i} - 26 \underset{i = 1}{\sum^{m}} i - \underset{i = 1}{\sum^{m}} 1]

The last sum is direct formula .