Evaluate- 1 1 2 - 1 1 3 - 1 1 - 4 - - - 1 1 19 -

S= 3 2 + 4 3 + 5 4 +.... 20 19 T _ n = n+2 n+1 =1+ 1 n+1 ∑ T _ n =∑ 1 +∑ 1 n+1 #160; #160; #160; #160; #160; #160; ...

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Standard XII

Mathematics

Summation by Sigma Method

Evaluate: 1 ...

Question

Evaluate:
$1 \frac{1}{2} + 1 \frac{1}{3} + 1 \frac{1}{4} + \dots + 1 \frac{1}{19} =$

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Solution

$S = \frac{3}{2} + \frac{4}{3} + \frac{5}{4} + . . . . \frac{20}{19}$
$T_{n} = \frac{n + 2}{n + 1} = 1 + \frac{1}{n + 1}$
$\sum T_{n} = \sum 1 + \sum \frac{1}{n + 1}$
$= n + (\frac{1}{2} + \frac{1}{3} + . . . . \frac{1}{19})$
$= 18 + \int_{1}^{18} \frac{d x}{x + 1}$
$= 18 + {(l n (x + 1))}_{1}^{18}$
$= 18 + l n 19 - l n 2$
$= 20.25$

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Evaluate:112+113+114+…+1119=

S=32+43+54+....2019Tn=n+2n+1=1+1n+1∑Tn=∑1+∑1n+1 =n+(12+13+....119) =18+∫181dxx+1 =18+(ln(x+1))181 =18+ln19−ln2 =20.25

Evaluate:
$1 \frac{1}{2} + 1 \frac{1}{3} + 1 \frac{1}{4} + \dots + 1 \frac{1}{19} =$