(1):: Σ_(i=1) ^n Σ_(j=1) ^n ∣i−j∣=? (2):: Σ_(i=1) ^n Σ_(j=i) ^n (1/j)=? (3):: Σ


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Question Number 147576 by qaz last updated on 22/Jul/21

$(1) :: \underset{i = 1}{\sum^{n}} \underset{j = 1}{\sum^{n}} ∣ i - j ∣= ?$ $(2) :: \underset{i = 1}{\sum^{n}} \underset{j = i}{\sum^{n}} \frac{1}{j} = ?$ $(3) :: \underset{i = 1}{\sum^{n^{2}}} [\sqrt{i}] = ?$
	Answered by gsk2684 last updated on 22/Jul/21
	$i)Σ_(i=1) ^n Σ_(j=1) ^n ∣i−j∣= 0+1+2+3+...+(n−1) +1+0+1+2+...+(n−2) +2+1+0+1+...+(n−3) +...... +(n−2)+(n−3)+...+1 +(n−1)+(n−2)+...+0 =n(0)+2(n−1)(1)+2(n−2)(2)+...2(1)(n−1) =Σ_(j=0) ^(n−1) 2j(n−j) =2{nΣ_(j=0) ^(n−1) j−Σ_(j=0) ^(n−1) j^2 } =2{n(((n−1)(n−1+1))/2)−(((n−1)(n−1+1)(2(n−1)+1))/6)} =2{((n^2 (n−1))/2)−((n(n−1)(2n−1))/6)} =2((n(n−1))/6){3n−2n+1} =((n(n−1)(n+1))/3) Σ_(i=1) ^n Σ_(j=1) ^n ∣i−j∣=((n(n^2 −1))/3) ii)Σ_(i=1) ^n Σ_(j=i) ^n (1/j)=(1/1)+(1/2)+(1/3)+...+(1/n) +(1/2)+(1/3)+...+(1/n) +(1/3)+...+(1/n) +........ +(1/n) =1+1+1+...+1 Σ_(i=1) ^n Σ_(j=i) ^n (1/j)=n iii)Σ_(i=1) ^n^2 [(√i)]=Σ_(i=1^2 ) ^(2^2 −1) 1+Σ_(i=2^2 ) ^(3^2 −1) 2+Σ_(i=3^2 ) ^(4^2 −1) 3+...+Σ_(i=(n−1)^2 ) ^(n^2 −1) (n−1)+[(√n^2 )] =3(1)+5(2)+7(3)+...+(2n−1)(n−1)+n =Σ_(j=1) ^(n−1) j(2j+1)+n =Σ_(j=1) ^(n−1) (2j^2 +j)+n =2Σ_(j=1) ^(n−1) j^2 +Σ_(j=1) ^(n−1) j+n =2(((n−1)(n−1+1)(2(n−1)+1))/6)+(((n−1)(n−1+1))/2)+n =((n(n−1)(2n−1))/3)+((n(n−1))/2)+n =n{(((n−1)(2n−1))/3)+(((n−1))/2)+1} =(n/6){2(2n^2 −3n+1)+3(n−1)+6} =(n/6){4n^2 −6n+2+3n−3+6} =(n/6){4n^2 −3n+5} Σ_(i=1) ^n^2 [(√i)]=((n(4n^2 −3n+5))/6)$
	$i) \underset{i = 1}{\sum^{n}} \underset{j = 1}{\sum^{n}} ∣ i - j ∣= 0 + 1 + 2 + 3 + . . . + (n - 1)$ $+ 1 + 0 + 1 + 2 + . . . + (n - 2)$ $+ 2 + 1 + 0 + 1 + . . . + (n - 3)$ $+ . . . . . .$ $+ (n - 2) + (n - 3) + . . . + 1$ $+ (n - 1) + (n - 2) + . . . + 0$ $= n (0) + 2 (n - 1) (1) + 2 (n - 2) (2) + . . . 2 (1) (n - 1)$ $= \underset{j = 0}{\sum^{n - 1}} 2 j (n - j)$ $= 2 {n \underset{j = 0}{\sum^{n - 1}} j - \underset{j = 0}{\sum^{n - 1}} j^{2}}$ $= 2 {n \frac{(n - 1) (n - 1 + 1)}{2} - \frac{(n - 1) (n - 1 + 1) (2 (n - 1) + 1)}{6}}$ $= 2 {\frac{n^{2} (n - 1)}{2} - \frac{n (n - 1) (2 n - 1)}{6}}$ $= 2 \frac{n (n - 1)}{6} {3 n - 2 n + 1}$ $= \frac{n (n - 1) (n + 1)}{3}$ $\underset{i = 1}{\sum^{n}} \underset{j = 1}{\sum^{n}} ∣ i - j ∣= \frac{n (n^{2} - 1)}{3}$ $i i) \underset{i = 1}{\sum^{n}} \underset{j = i}{\sum^{n}} \frac{1}{j} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{n}$ $+ \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{n}$ $+ \frac{1}{3} + . . . + \frac{1}{n}$ $+ . . . . . . . .$ $+ \frac{1}{n}$ $= 1 + 1 + 1 + . . . + 1$ $\underset{i = 1}{\sum^{n}} \underset{j = i}{\sum^{n}} \frac{1}{j} = n$ $i i i) \underset{i = 1}{\sum^{n^{2}}} [\sqrt{i}] = \underset{i = 1^{2}}{\sum^{2^{2} - 1}} 1 + \underset{i = 2^{2}}{\sum^{3^{2} - 1}} 2 + \underset{i = 3^{2}}{\sum^{4^{2} - 1}} 3 + . . . + \underset{i = {(n - 1)}^{2}}{\sum^{n^{2} - 1}} (n - 1) + [\sqrt{n^{2}}]$ $= 3 (1) + 5 (2) + 7 (3) + . . . + (2 n - 1) (n - 1) + n$ $= \underset{j = 1}{\sum^{n - 1}} j (2 j + 1) + n$ $= \underset{j = 1}{\sum^{n - 1}} (2 j^{2} + j) + n$ $= 2 \underset{j = 1}{\sum^{n - 1}} j^{2} + \underset{j = 1}{\sum^{n - 1}} j + n$ $= 2 \frac{(n - 1) (n - 1 + 1) (2 (n - 1) + 1)}{6} + \frac{(n - 1) (n - 1 + 1)}{2} + n$ $= \frac{n (n - 1) (2 n - 1)}{3} + \frac{n (n - 1)}{2} + n$ $= n {\frac{(n - 1) (2 n - 1)}{3} + \frac{(n - 1)}{2} + 1}$ $= \frac{n}{6} {2 (2 n^{2} - 3 n + 1) + 3 (n - 1) + 6}$ $= \frac{n}{6} {4 n^{2} - 6 n + 2 + 3 n - 3 + 6}$ $= \frac{n}{6} {4 n^{2} - 3 n + 5}$ $\underset{i = 1}{\sum^{n^{2}}} [\sqrt{i}] = \frac{n (4 n^{2} - 3 n + 5)}{6}$
	Commented by qaz last updated on 22/Jul/21

	$thanks a lot sir$
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