1-i-1-n-j-1-n-i-j-2-i-1-n-j-i-n-1-j-3-i-1-n-2-i-

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) \underset{i = 1}{\sum^{n}} \underset{j = 1}{\sum^{n}} ∣ i - j ∣= 0 + 1 + 2 + 3 + \dots + (n - 1)

+ 1 + 0 + 1 + 2 + \dots + (n - 2)

+ 2 + 1 + 0 + 1 + \dots + (n - 3)

+ \dots \dots

+ (n - 2) + (n - 3) + \dots + 1

+ (n - 1) + (n - 2) + \dots + 0

= n (0) + 2 (n - 1) (1) + 2 (n - 2) (2) + \dots 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} + \dots + \frac{1}{n}

+ \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}

+ \frac{1}{3} + \dots + \frac{1}{n}

+ \dots \dots . .

+ \frac{1}{n}

= 1 + 1 + 1 + \dots + 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 + \dots + \underset{i = {(n - 1)}^{2}}{\sum^{n^{2} - 1}} (n - 1) + [\sqrt{n^{2}}]

= 3 (1) + 5 (2) + 7 (3) + \dots + (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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