Sum of the series S -1-1-21-2-1-31-2-3-1-41-2-3-4-upto 19 terms isA- 116B- 118C- 110D- 98E- 104

The correct option is E 104 n th term is of the form 1/n×∑ n The series is of the term 2/2+3/2+4/2+5/2.....20/2 Thus, the sum is 1/2×∑ n 1=1/2×(20× 21/2) 1=1 ...

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Series

Sum of the se...

Question

Sum of the series
$S = 1 + \frac{1}{2} (1 + 2) + \frac{1}{3} (1 + 2 + 3) + \frac{1}{4} (1 + 2 + 3 + 4) +$
.....upto 19 terms is

110

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104

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116

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118

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Solution

The correct option is C
104

n th term is of the form $\frac{1}{n} \times \sum n$

The series is of the term $\frac{2}{2} + \frac{3}{2} + \frac{4}{2} + \frac{5}{2} . . . . . \frac{20}{2}$

Thus, the sum is $\frac{1}{2} \times \sum n - 1 = \frac{1}{2} \times (\frac{20 \times 21}{2}) - 1 = 104.$

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Sum of the series S=1+12(1+2)+13(1+2+3)+14(1+2+3+4)+ .....upto 19 terms is

The correct option is C 104 n th term is of the form 1n×∑n The series is of the term 22+32+42+52.....202 Thus, the sum is 12×∑n−1=12×(20×212)−1=104.

Sum of the series
$S = 1 + \frac{1}{2} (1 + 2) + \frac{1}{3} (1 + 2 + 3) + \frac{1}{4} (1 + 2 + 3 + 4) +$
.....upto 19 terms is

The correct option is C
104

n th term is of the form $\frac{1}{n} \times \sum n$

The series is of the term $\frac{2}{2} + \frac{3}{2} + \frac{4}{2} + \frac{5}{2} . . . . . \frac{20}{2}$

Thus, the sum is $\frac{1}{2} \times \sum n - 1 = \frac{1}{2} \times (\frac{20 \times 21}{2}) - 1 = 104.$