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 [ Σ_(i=1) ^(100) (a


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

$L e t a_{1}, a_{2}, a_{3}, . . ., a_{n} b e r e a l n u m b e r s s u c h t h a t :$ $\sqrt{a_{1}} + \sqrt{a_{2} - 1} + \sqrt{a_{3} - 2} + . . . + \sqrt{a_{n} - (n - 1)} = \frac{1}{2} (a_{1} + a_{2} + a_{3} + . . . + a_{n}) - \frac{n (n - 3)}{4}$ $T h e n f i n d t h e v a l u e o f [\underset{i = 1}{\sum^{100}} (a_{i})] .$
	Answered by witcher3 last updated on 14/May/23
	$(√a)≤(1/2)(a)+(1/2),∀a≥0 ⇒Σ_(k=1) ^n (√(a_k −(k−1)))≤Σ_(k=1) ^n (1/2)(a_k −k+2)=(1/2)(a_1 +....+a_n )−(1/2)(((n(n+1))/2)−((4n)/2)) ⇒≤(1/2)(a_1 +....+a_n )−((n(n−3))/4) withe Equality ⇔ it (√(a_k −(k−1)))=1⇒a_k =k,∀k∈{1.....n} ⇒Σ_(k=1) ^(100) a_k =50(101)=5050$
	$\sqrt{a} ⩽ \frac{1}{2} (a) + \frac{1}{2}, \forall a ⩾ 0$ $\Rightarrow \underset{k = 1}{\sum^{n}} \sqrt{a_{k} - (k - 1)} ⩽ \underset{k = 1}{\sum^{n}} \frac{1}{2} (a_{k} - k + 2) = \frac{1}{2} (a_{1} + . . . . + a_{n}) - \frac{1}{2} (\frac{n (n + 1)}{2} - \frac{4 n}{2})$ $\Rightarrow⩽ \frac{1}{2} (a_{1} + . . . . + a_{n}) - \frac{n (n - 3)}{4}$ $withe Equality \Leftrightarrow$ $it \sqrt{a_{k} - (k - 1)} = 1 \Rightarrow a_{k} = k, \forall k \in {1 . . . . . n}$ $\Rightarrow \underset{k = 1}{\sum^{100}} a_{k} = 50 (101) = 5050$
	Commented by York12 last updated on 14/May/23
	$let b_i =(√(a_(i ) −(i−1) )) → a_i =(b_i )^2 +i−1 ∴Σ_(j=1 ) ^n [b_(i ) ]= (1/2)(a_1 +a_2 +a_3 +...+a_n _(Σ_(i=1) ^n [a_(i ) ]) )−((n(n−3))/4) =(1/2)(Σ_(i=1) ^n [(b_i )^2 ]+Σ_(i=1) ^n [i]−Σ_(i=1) ^n (1))−((n(n−3))/4) =(1/2) Σ_(i=1) ^n [(b_i )^2 ]+(n/2) ∴ 2Σ_(i=1) ^n (b_i )=Σ_(i=1) ^n [(b_i )^2 ]+n → Σ_(i=1) ^n [(b_(i ) )^2 −2b_i +1]=0 ∴ [b_i −1]^2 =0 → b_i =1 ∴ a_i =i → Σ_(i=1) ^(100) (a_i )=Σ_(i=1) ^(100) (i)=((100×101)/2)=5050 (That′s it) {By York.W}$
	$l e t b_{i} = \sqrt{a_{i} - (i - 1)} \to a_{i} = {(b_{i})}^{2} + i - 1$ $∴ \underset{j = 1}{\sum^{n}} [b_{i}] = \frac{1}{2} (\underset{\underset{i = 1}{\sum^{n}} [a_{i}]}{\underset{⏟}{a_{1} + a_{2} + a_{3} + . . . + a_{n}}}) - \frac{n (n - 3)}{4}$ $= \frac{1}{2} (\underset{i = 1}{\sum^{n}} [{(b_{i})}^{2}] + \underset{i = 1}{\sum^{n}} [i] - \underset{i = 1}{\sum^{n}} (1)) - \frac{n (n - 3)}{4}$ $= \frac{1}{2} \underset{i = 1}{\sum^{n}} [{(b_{i})}^{2}] + \frac{n}{2}$ $∴ 2 \underset{i = 1}{\sum^{n}} (b_{i}) = \underset{i = 1}{\sum^{n}} [{(b_{i})}^{2}] + n \to \underset{i = 1}{\sum^{n}} [{(b_{i})}^{2} - 2 b_{i} + 1] = 0$ $∴ {[b_{i} - 1]}^{2} = 0 \to b_{i} = 1$ $∴ a_{i} = i \to \underset{i = 1}{\sum^{100}} (a_{i}) = \underset{i = 1}{\sum^{100}} (i) = \frac{100 \times 101}{2} = 5050 ({T h a t}^{'} s i t)$ ${B y Y o r k . W}$
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