If-a-n-a-n-1-1-for-every-positive-integer-greater-than-1-then-a-1-a-2-a-3-a-100-equals-1-5000-a-1-2-5050-a-1-3-5051-a-1-3-5052-a-2-

Question Number 25085 by Tinkutara last updated on 03/Dec/17

I f a_{n} - a_{n - 1} = 1 f o r e v e r y p o s i t i v e

i n t e g e r g r e a t e r t h a n 1, t h e n a_{1} + a_{2} + a_{3}

+ \dots a_{100} e q u a l s

(1) 5000 . a_{1}

(2) 5050 . a_{1}

(3) 5051 . a_{1}

(3) 5052 . a_{2}

Answered by ajfour last updated on 03/Dec/17

\underset{r = 1}{\sum^{100}} a_{r} = a_{1} + (a_{1} + 1) + (a_{2} + 1) + \dots .

\dots + (a_{99} + 1)

= a_{1} + (a_{1} + 1) + (a_{1} + 2) + (a_{2} + 3) + . .

\dots . + (a_{1} + 99)

= 100 a_{1} + \frac{99 \times 100}{2}

= 100 (a_{1} + \frac{99}{2}) .

b u t i f i t w e r e g i v e n t h a t

a_{n} - a_{n - 1} = a_{1}

\Rightarrow a_{2} = 2 a_{1}, a_{3} = 3 a_{1}, \dots . a n d s o o n

t h e n \underset{r = 1}{\sum^{100}} a_{r} = a_{1} (1 + 2 + 3 + \dots 100)

= (\frac{100 \times 101}{2}) a_{1} = 5050 a_{1} .

o p t i o n (2) .

Commented by Tinkutara last updated on 03/Dec/17

Thank you Sir!