Let-a-1-a-2-a-3-a-n-be-real-numbers-such-that-a-1-a-2-1-a-3-2-a-n-n-1-1-2-a-1-a-2-a-3-a-n-n-n-3-4-Then-find-the-value-of-i-1-10

Question Number 192277 by York12 last updated on 13/May/23

L e t a_{1}, a_{2}, a_{3}, \dots, a_{n} b e r e a l n u m b e r s s u c h t h a t :

\sqrt{a_{1}} + \sqrt{a_{2} - 1} + \sqrt{a_{3} - 2} + \dots + \sqrt{a_{n} - (n - 1)} = \frac{1}{2} (a_{1} + a_{2} + a_{3} + \dots + a_{n}) - \frac{n (n - 3)}{4}

T h e n f i n d t h e v a l u e o f [\underset{i = 1}{\sum^{100}} (a_{i})] .

Answered by witcher3 last updated on 14/May/23

\sqrt{a} ⩽ \frac{1}{2} (a) + \frac{1}{2}, \forall a ⩾ 0

\Rightarrow \underset{k = 1}{\sum^{n}} \sqrt{a_{k} - (k - 1)} ⩽ \underset{k = 1}{\sum^{n}} \frac{1}{2} (a_{k} - k + 2) = \frac{1}{2} (a_{1} + \dots . + a_{n}) - \frac{1}{2} (\frac{n (n + 1)}{2} - \frac{4 n}{2})

\Rightarrow⩽ \frac{1}{2} (a_{1} + \dots . + a_{n}) - \frac{n (n - 3)}{4}

withe Equality \Leftrightarrow

it \sqrt{a_{k} - (k - 1)} = 1 \Rightarrow a_{k} = k, \forall k \in {1 \dots . . n}

\Rightarrow \underset{k = 1}{\sum^{100}} a_{k} = 50 (101) = 5050

Commented by York12 last updated on 14/May/23

l e t b_{i} = \sqrt{a_{i} - (i - 1)} \to a_{i} = {(b_{i})}^{2} + i - 1

∴ \underset{j = 1}{\sum^{n}} [b_{i}] = \frac{1}{2} (\underset{\underset{i = 1}{\sum^{n}} [a_{i}]}{\underset{⏟}{a_{1} + a_{2} + a_{3} + \dots + a_{n}}}) - \frac{n (n - 3)}{4}

= \frac{1}{2} (\underset{i = 1}{\sum^{n}} [{(b_{i})}^{2}] + \underset{i = 1}{\sum^{n}} [i] - \underset{i = 1}{\sum^{n}} (1)) - \frac{n (n - 3)}{4}

= \frac{1}{2} \underset{i = 1}{\sum^{n}} [{(b_{i})}^{2}] + \frac{n}{2}

∴ 2 \underset{i = 1}{\sum^{n}} (b_{i}) = \underset{i = 1}{\sum^{n}} [{(b_{i})}^{2}] + n \to \underset{i = 1}{\sum^{n}} [{(b_{i})}^{2} - 2 b_{i} + 1] = 0

∴ {[b_{i} - 1]}^{2} = 0 \to b_{i} = 1

∴ a_{i} = i \to \underset{i = 1}{\sum^{100}} (a_{i}) = \underset{i = 1}{\sum^{100}} (i) = \frac{100 \times 101}{2} = 5050 ({T h a t}^{'} s i t)

{B y Y o r k . W}

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