Evaluate: $1 + 3 + 5 + 7 + 9 + 11 + 13$

25

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36

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49

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100

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Solution

The correct option is B $49$
Sum of first $n$ consecutive odd numbers is $n^{2}$
Here, $n = 7$
Therefore sum is $7^{2} = 49$

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Similar questions

Q. Evaluate:

(i) \frac{3}{7} + \frac{- 4}{9} - \frac{- 11}{7} - \frac{7}{9}

(i i) \frac{2}{3} + \frac{- 4}{5} - \frac{1}{3} - \frac{2}{5}

(i i i) \frac{4}{7} - \frac{- 8}{9} - \frac{- 13}{7} + \frac{17}{9}

Evaluate each of the following:

$(a) (- 30) \div 10$

$(b) 50 \div (- 5)$

$(c) (- 36) \div (- 9)$

$(d) (- 49) \div 49$

$(e) 13 \div [- 2 + 1]$

$(f) 0 \div (- 12)$

$(g) (- 31) \div [(- 30) + (- 1)]$

$(h) [(- 36) \div 12] \div 3$

$(i) [- 6 + 5] \div [- 2 + 1]$

Evaluate :

$(i) \frac{3}{7} + \frac{- 4}{9} + \frac{- 11}{7} + \frac{7}{9} (i i) \frac{2}{3} + \frac{- 4}{5} + \frac{1}{3} + \frac{2}{5} (i i i) \frac{4}{7} + 0 + \frac{- 8}{9} + \frac{- 13}{7} + \frac{17}{9} (i v) \frac{3}{8} + \frac{- 5}{12} + \frac{3}{7} + \frac{3}{12} + \frac{- 5}{8} + \frac{- 2}{7}$

Q. Without adding, find the sum:
(i) (1 + 3 + 5 + 7 + 9 + 11 + 13)
(ii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19)
(iii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23)

Q. If

9 \times 3 = 36

;

11 \times 7 = 81

, then what is the value of

5 \times 13

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Evaluate: 1+3+5+7+9+11+13

The correct option is B 49Sum of first n consecutive odd numbers is n2 Here, n=7Therefore sum is 72=49

Evaluate: $1 + 3 + 5 + 7 + 9 + 11 + 13$

The correct option is B $49$
Sum of first $n$ consecutive odd numbers is $n^{2}$
Here, $n = 7$
Therefore sum is $7^{2} = 49$