Question

Sum of the series S=1+=(1+2)+
=1+( 1+2)+{(1+2+3)+L(1+2+3+4) + ... upto 20 tems is​

Answer 1

Step-by-step explanation:

1+(1+2)/2+(1+2+3)/3+(1+2+3+4)/4+…(1+2+3+…20)/20

=1+(1+1/2)+2+(2+1/2)+………10+(10+1/2)

=(1+1)+(2+2)+(3+3)+…….(10+10)+10×1/2

=2(1+2+3+4+……10)+10×1/2

=2×10/2×(1+10)+5

=10×11+5=110+5=115 ans.

Answer 2

Answer :

›»› The sum of the series = 115

Given :

• $\sf{S = 1 + \dfrac{1}{2}(1 + 2) + \dfrac{1}{3}(1 + 2 + 3) + \dfrac{1}{4}(1 + 2 + 3 + 4)}$

To Find :

• Sum of the series.

Required Solution :

$\hookrightarrow$$\tt{S = 1 + \dfrac{1}{2}(1 + 2) + \dfrac{1}{3}(1 + 2 + 3) + \dfrac{1}{4}(1 + 2 + 3 + 4)}$

$\hookrightarrow$$\tt{S = 1 + \dfrac{1}{2} \times 3 + \dfrac{1}{3} \times 6 + \dfrac{1}{4} \times 10}$

$\hookrightarrow$$\bf{S = 1 + \dfrac{3}{2}+ \dfrac{6}{3} + \dfrac{10}{4}}$

Now ,

→ First term = 1

→ Common difference = a₂ - a₁

→ Common difference = 6/3 - 3/2

→ Common difference = 2/1 - 3/2

→ Common difference = 1/2

Now, we have First term, common difference and nth term of the series,

• First term = 1
• Common difference = ½
• nth term = 20

And we need to find Sum of the series.

We can find Sum of the series by using the formula :

$\sf{S_{n} = \dfrac{n}{2} \bigg[2a + (n - 1)d \bigg]}$

Here,

• S is the Sum.
• n is the nth term.
• a is the First term.
• d is the Common difference.

So let's find Sum of the series !

$\tt{: \implies S_{n} = \dfrac{n}{2} \bigg[2a + (n - 1)d \bigg]}$

$\tt{: \implies S_{20} = \cancel{\dfrac{20}{2}} \bigg[2 \times 1 + (20 - 1) \dfrac{1}{2} \bigg]}$

$\tt{: \implies S_{20} = 10\bigg[2 \times 1 + (20 - 1) \dfrac{1}{2} \bigg]}$

$\tt{: \implies S_{20} = 10 \bigg[2 + 19 \times \dfrac{1}{2} \bigg]}$

$\tt{: \implies S_{20} = 10 \bigg[2 + \cancel{\dfrac{19}{2}} \bigg]}$

$\tt{: \implies S_{20} = 10 \big[2 + 9.5 \big]}$

$\tt{: \implies S_{20} = 10 \times 11.5}$

$\bf{: \implies \underline{ \: \: \underline{ \red{ \: \: S_{20} = 115 \: \: }} \: \: }}$

║Hence, the Sum of the series is 115.║