Let A = ∑ _ i = 1 ^ 10 ∑ _ j = 1 ^ 10 min\ i, j\ and B = ∑ _ i = 1 ^ 10 ∑ _ j = 1 ^ 10 max\ i, j\ . Then A + B is equal to _____.

Correct answer: 0 A = ∑ _ i = 1 ^ 10 ∑ _ j = 1 ^ 10 min \ i, j\ B = ∑ _ i = 1 ^ 10 ∑ _ j = 1 ^ 10 max \ i, j\ A = ∑ _ j = 1 ^ 10 min(i, 1) + min(j, 2) + min (i, 10) = (1 + 1 + 1 + + 1) _ ⏟ _ 19times + (2 + 2 + 2 + 2) _ ⏟ _ 17times + (3 + 3 + 3 + 3) _ ⏟ _ 15times + . . . B = ∑ _ j = 1 ^ 10 max (i, 1) + max(j, 2) + max (i, 10) = (10 + 10 + + 10) _ ⏟ _ 19times + (9 + 9 + + 9) _ ⏟ _ 17times + . . . i.e. each element of the ordered pair \ i, j\ will appear either in Aor B So A + B will be sum of all elements of the ordered pair A + B = 20(1 + 2 + 3 + + 10) = 20 × (10 × 11)/(2) =10 × 110 = 1100

JEE Main 202226 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 26 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2022 · 26 June · Morning Shift · Mathematics · Sequences and Series

Let A = ∑ _ i = 1 ^ 10 ∑ _ j = 1 ^ 10 min\ i, j\ and B = ∑ _ i = 1...

Let

A = \mathop{∑}_{i = 1}^{10}\mathop{∑}_{j = 1}^{10}min\left\{i, j\right\}

and

B = \mathop{∑}_{i = 1}^{10}\mathop{∑}_{j = 1}^{10}max\left\{i, j\right\}

. Then

A + B

is equal to _____.

Correct answer

Step-by-step solution

A = \mathop{∑}_{i = 1}^{10}\mathop{∑}_{j = 1}^{10}min $\left\{i, j\right\}$

$B = \mathop{∑}_{i = 1}^{10}\mathop{∑}_{j = 1}^{10}max \left\{i, j\right\}A = \mathop{∑}_{j = 1}^{10}min\left(i, 1\right) + min\left(j, 2\right) + \ldots min \left(i, 10\right)$ = \mathop{\mathop{\left(1 + 1 + 1 + \ldots + 1\right)}_{⏟}}_{19times} + \mathop{\mathop{\left(2 + 2 + 2 \ldots + 2\right) $}_{⏟}}_{17times} + \mathop{\mathop{\left(3 + 3 + 3 \ldots + 3\right)}_{⏟}}_{15times} + . . .$ B = \mathop{∑}_{j = 1}^{10}max \left(i, 1\right) + max\left(j, 2\right) + \ldots max \left(i, 10\right)

$= \mathop{\mathop{\left(10 + 10 + \ldots + 10\right)}_{⏟}}$ _{19times} + \mathop{\mathop{\left(9 + 9 + \ldots + 9\right) $}_{⏟}}_{17times} + . . .i.e. each element of$ the ordered pair \left\{i, j\right\} will appear either in A\text{or} $B$

So A + B $will be sum of all elements of the ordered pair A + B = 20\left(1 + 2 + 3 + \ldots + 10\right)$

$= 20 \times \dfrac{10 \times 11}{2} =10 \times 110 = 1100$

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