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Class 11 RD Sharma Solutions – Chapter 21 Some Special Series- Exercise 21.2

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Sum the following series to n terms:

Question 1. 3 + 5 + 9 + 15 + 23 + . . . . n terms

Solution:

We are given the series: 3 + 5 + 9 + 15 + 23 + . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 3 + 5 + 9 + 15 + 23 + . . . . + an–1 + an  . . . .(1)

S = 3 + 5 + 9 + 15 + 23 + . . . . + an–2 + an–1 + an  . . . .(2)

On subtracting (2) from (1), we get

=> S – S = [3 + (5 + 9 + 15 + 23 + . . . . + an–1 + an)] – [(3 + 5 + 9 + 15 + 23 + . . . . + an–2 + an–1) +an]

=> 0 = 3 + [(5 – 3) + (9 – 5) + (15 – 9) + . . . . + (an – an–1)] – an

=> an = 3 + [2 + 4 + 6 + . . . . (n–1) terms]

As the series 2+ 4 + 6 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 4–2 = 2. So, we get,

=> an = 3 + (n–1) [2(2)+(n–2)2]/2      

= 3+ (n–1) [4+2n–4]/2

= 3 + n(n–1)

= n2 – n+3

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k + \sum_{k=1}^{n} 3

 \frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+3n

n\left[\frac{(n+1)(2n+1)-3(n+1)+18}{6}\right]

\frac{n}{6}\left[2n^2+n+2n+1-3n-3+18\right]

\frac{n}{6}\left[2n^2+16\right]

 \frac{n(n^2+8)}{3}

Therefore, sum of the given series up to n terms is  \frac{n(n^2+8)}{3}.

Question 2. 2 + 5 + 10 + 17 + 26 + . . . . n terms

Solution:

We are given the series: 2 + 5 + 10 + 17 + 26 + . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 2 + 5 + 10 + 17 + 26 + . . . . + an–1 + an  . . . .(1)

S = 2 + 5 + 10 + 17 + 26 + . . . . + an–2 + an–1 + an  . . . .(2)

On subtracting (2) from (1), we get

=> S – S = [2 + (5 + 10 + 17 + 26 + . . . . + an–1 + an)] – [(2 + 5 + 10 + 17 + 26 + . . . . + an–2 + an–1) +an]

=> 0 = 2 + [(5 – 2) + (10 – 5) + (17 – 10) + . . . . + (an – an–1)] – an

=> an = 2 + [3 + 5 + 7 + . . . . (n–1) terms]

As the series 3 + 5 + 7 + . . . . (n–1) terms is an A.P., with first term(a) = 3 and common difference(d) = 5–3 = 2. So, we get,

=> an = 2 + (n–1) [2(3)+(n–2)2]/2      

= 2+ (n–1) [6+2n–4]/2

= 2 + (n–1)(n+1)

= n2 – 1 + 2

= n2 + 1

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} 1

 \frac{n(n+1)(2n+1)}{6}+n

 \frac{n(n+1)(2n+1)+6n}{6}

 \frac{n(2n^2+3n+7)}{6}

Therefore, sum of the given series up to n terms is  \frac{n(2n^2+3n+7)}{6}.

Question 3. 1 + 3 + 7 + 13 + 21 + 31 +  . . . . n terms

Solution:

We are given the series: 1 + 3 + 7 + 13 + 21 + 31 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 1 + 3 + 7 + 13 + 21 + 31 +  . . . . + an–1 + an  . . . .(1)

S = 1 + 3 + 7 + 13 + 21 + 31 +  . . . . + an–2 + an–1 + an  . . . .(2)

On subtracting (2) from (1), we get

=> S – S = [1 + (3 + 7 + 13 + 21 + 31 +  . . . . + an–1 + an)] – [(1 + 3 + 7 + 13 + 21 + 31 +  . . . . + an–2 + an–1) + an]

=> 0 = 1 + [(3 – 1) + (7 – 3) + (13 – 7) + . . . . + (an – an–1)] – an

=> an = 1 + [2 + 4 + 6 + . . . . (n–1) terms]

As the series 2 + 4 + 6 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 4–2 = 2. So, we get,

=> an = 1 + (n–1) [2(2)+(n–2)2]/2      

= 1+ (n–1) [4+2n–4]/2

= 1 + n(n–1)

= n2 – n+1

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1

\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+n

\frac{n(n+1)(2n+1)-3n(n+1)+6n}{6}

\frac{n(2n^2+n+2n+1-3n-3+6)}{6}

\frac{n(2n^2+4)}{6}

\frac{n(n^2+2)}{3}

Therefore, sum of the given series up to n terms is \frac{n(n^2+2)}{3}.

Question 4. 3 + 7 + 14 + 24 + 37 +  . . . . n terms 

Solution: 

We are given the series: 3 + 7 + 14 + 24 + 37 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 3 + 7 + 14 + 24 + 37 +  . . . . + an–1 + a . . . .(1)

S = 3 + 7 + 14 + 24 + 37 +  . . . . + an–2 + an–1 + an  . . . .(2)

On subtracting (2) from (1), we get

=> S – S = [3 + (7 + 14 + 24 + 37 +  . . . . + an–1 + an)] – [(3 + 7 + 14 + 24 + 37 +  . . . . + an–2 + an–1) + an]

=> 0 = 3 + [(7 – 3) + (14 – 7) + (24 – 14) + . . . . + (an – an–1)] – an

=> an = 3 + [4 + 7 + 10 + . . . . (n–1) terms]

As the series 4 + 7 + 10 + . . . . (n–1) terms is an A.P., with first term(a) = 4 and common difference(d) = 7–4 = 3. So, we get,

=> an = 3 + (n–1) [2(4)+(n–2)3]/2      

= 3+ (n–1) [8+3n–6]/2

= 3 + (n–1)(3n+2)/2

\frac{6+(n-1)(3n+2)}{2}

\frac{6+3n^2-n-2}{2}

\frac{3n^2-n+4}{2}

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{n} a_k = \frac{1}{2}\left[\sum_{k=1}^{n} 3k^2 - \sum_{k=1}^{n} k + \sum_{k=1}^{n} 4\right]

\frac{3}{2}\sum_{k=1}^{n} k^2 - \frac{1}{2}\sum_{k=1}^{n} k + \sum_{k=1}^{n} 2

\frac{3}{2}[\frac{n(n+1)(2n+1)}{6}]-\frac{1}{2}[\frac{n(n+1)}{2}]+2n

\frac{n(n+1)(2n+1)-n(n+1)+8n}{4}

\frac{n(2n^2+2n+8)}{4}

\frac{n(n^2+n+4)}{2}

Therefore, sum of the given series up to n terms is \frac{n(n^2+n+4)}{2} .

Question 5. 1 + 3 + 6 + 10 + 15 +  . . . . n terms

Solution:

We are given the series: 1 + 3 + 6 + 10 + 15 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 1 + 3 + 6 + 10 + 15 +  . . . . + an–1 + an  . . . .(1)

S = 1 + 3 + 6 + 10 + 15 +  . . . . + an–2 + an–1 + a . . . .(2)

On subtracting (2) from (1), we get

=> S – S = [1 + (3 + 6 + 10 + 15 +  . . . . + an–1 + an)] – [(1 + 3 + 6 + 10 + 15 +  . . . . + an–2 + an–1) + an]

=> 0 = 1 + [(3 – 1) + (6 – 3) + (10 – 6) + . . . . + (an – an–1)] – an

=> an = 1 + [2 + 3 + 4 + . . . . (n–1) terms]

As the series 2 + 3 + 4 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 3–2 = 1. So, we get,

=> an = 1 + (n–1) [2(2)+(n–2)1]/2      

= 1+ (n–1) [4+n–2]/2

= 1 + (n–1)(n+2)/2

\frac{2+(n-1)(n+2)}{2}

\frac{2+n^2+2n-n-2}{2}

\frac{n^2+n}{2}

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{n} a_k = \frac{1}{2}\left[\sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k\right]

\frac{1}{2}\sum_{k=1}^{n} k^2 + \frac{1}{2}\sum_{k=1}^{n} k

\frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6}\right] + \frac{1}{2}\left[\frac{n(n+1)}{2}\right]

\frac{n(n+1)}{2}\left[\frac{2n+1}{6}+\frac{1}{2}\right]

\frac{n(n+1)(2n+4)}{12}

\frac{n(n+1)(n+2)}{6}

Therefore, sum of the given series up to n terms is \frac{n(n+1)(n+2)}{6} .

Question 6. 1 + 4 + 13 + 40 + 121 +  . . . . n terms 

Solution:

We are given the series: 1 + 4 + 13 + 40 + 121 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 1 + 4 + 13 + 40 + 121 +  . . . . + an–1 + an  . . . .(1)

S = 1 + 4 + 13 + 40 + 121 +  . . . . + an–2 + an–1 + an  . . . .(2)

On subtracting (2) from (1), we get

=> S – S = [1 + (4 + 13 + 40 + 121 +  . . . . + an–1 + an)] – [(1 + 4 + 13 + 40 + 121 +  . . . . + an–2 + an–1) + an]

=> 0 = 1 + [(4 – 1) + (13 – 4) + (40 – 13) + . . . . + (an – an–1)] – an

=> an = 1 + [3 + 9 + 27 + . . . . (n–1) terms]

As the series 3 + 9 + 27 + . . . . (n–1) terms is a G.P., with first term(a) = 3 and common ratio(r) = 9/3 = 3. So, we get,

=> an = 1 + 3(3n-1–1)/(3–1) 

= 1+ 3(3n-1–1)/2 

= 1 + 3n/2 – 3/2

= 3n/2 – 1/2

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{n} a_k = \frac{1}{2}\sum_{k=1}^{n} 3^n - \frac{1}{2}\sum_{k=1}^{n} 1

\frac{3^1+3^2+3^3+...+3^n}{2}-\frac{n}{2}

\frac{3(3^n-1)}{(3-1)2}-\frac{n}{2}

\frac{3(3^n)-2n-3}{4}

\frac{3^{n+1}-2n-3}{4}

Therefore, sum of the given series up to n terms is \frac{3^{n+1}-2n-3}{4} .

Question 7. 4 + 6 + 9 + 13 + 18 +  . . . . n terms 

Solution:

We are given the series: 4 + 6 + 9 + 13 + 18 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 4 + 6 + 9 + 13 + 18 +  . . . . + an–1 + an  . . . .(1)

S = 4 + 6 + 9 + 13 + 18 +  . . . . + an–2 + an–1 + an  . . . .(2)

On subtracting (2) from (1), we get

=> S – S = [4 + (6 + 9 + 13 + 18 +  . . . . + an–1 + an)] – [(4 + 6 + 9 + 13 + 18 +  . . . . + an–2 + an–1) + an]

=> 0 = 4 + [(6 – 4) + (9 – 6) + (13 – 9) + . . . . + (an – an–1)] – an

=> an = 4 + [2 + 3 + 4 + . . . . (n–1) terms]

As the series 2 + 3 + 4 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 3–2 = 1. So, we get,

=> an = 4 + (n–1) [2(2)+(n–2)1]/2

= 4 + (n–1)(n+2)/2

\frac{8+(n-1)(n+2)}{2}

\frac{n^2+n+6}{2}

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{n} a_k = \frac{1}{2}\left[\sum_{k=1}^{n} n^2 + \sum_{k=1}^{n} n + \sum_{k=1}^{n} 6\right]

\frac{1}{2}\sum_{k=1}^{n} n^2 + \frac{1}{2}\sum_{k=1}^{n} n + \sum_{k=1}^{n} 3

\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}] + \frac{1}{2}[\frac{n(n+1)}{2}] + 3n

\frac{n[(n+1)(2n+1)+3(n+1)+36]}{12}

\frac{n[2n^2+6n+40]}{12}

\frac{n[n^2+3n+20]}{6}

Therefore, sum of the given series up to n terms is \frac{n[n^2+3n+20]}{6} .

Question 8. 2 + 4 + 7 + 11 + 16 +  . . . . n terms  

Solution:

We are given the series: 2 + 4 + 7 + 11 + 16 +  . . . . n terms.

Let’s take S as the sum of this series. Therefore,

S = 2 + 4 + 7 + 11 + 16 +  . . . . + an–1 + an  . . . .(1)

S = 2 + 4 + 7 + 11 + 16 +  . . . . + an–2 + an–1 + an  . . . .(2)

On subtracting (2) from (1), we get

=> S – S = [2 + (4 + 7 + 11 + 16 +  . . . . + an–1 + an)] – [(2 + 4 + 7 + 11 + 16 +  . . . . + an–2 + an–1) + an]

=> 0 = 2 + [(4 – 2) + (7 – 4) + (11 – 7) + . . . . + (an – an–1)] – an

=> an = 2 + [2 + 3 + 4 + . . . . (n–1) terms]

As the series 2 + 3 + 4 + . . . . (n–1) terms is an A.P., with first term(a) = 2 and common difference(d) = 3–2 = 1. So, we get,

=> an = 2 + (n–1) [2(2)+(n–2)1]/2

= 2 + (n–1)(n+2)/2

\frac{4+n^2+2n-n-2}{2}

\frac{n^2+n+2}{2}

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{n} a_k = \frac{1}{2}\left[\sum_{k=1}^{n} n^2 + \sum_{k=1}^{n} n + \sum_{k=1}^{n} 2\right]

\frac{1}{2}\sum_{k=1}^{n} n^2 + \frac{1}{2}\sum_{k=1}^{n} n + \sum_{k=1}^{n} 1

\frac{1}{2}[\frac{n(n+1)(2n+1)}{6}]+\frac{1}{2}[\frac{n(n+1)}{2}]+n

\frac{n[(n+1)(2n+1)+3(n+1)+12n]}{12}

\frac{n(2n^2+6n+16)}{12}

\frac{n(n^2+3n+8)}{6}

Therefore, sum of the given series up to n terms is \frac{n(n^2+3n+8)}{6} .

Question 9. \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....n\hspace{0.1cm}terms    

Solution: 

The nth term of the given series would be,

an\frac{1}{(3n-2)(3n+1)}

\frac{1}{3}\left[\frac{1}{3n-2}-\frac{1}{3n+1}\right]

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{n} a_k = \frac{1}{3}\sum_{k=1}^{n} \left[\frac{1}{3n-2}-\frac{1}{3n+1}\right]

\frac{1}{3}\left[(1-\frac{1}{4})+(\frac{1}{4}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{10})...(\frac{1}{3n-2}-\frac{1}{3n+1})\right]

\frac{1}{3}\left[1-\frac{1}{3n+1}\right]

\frac{3n}{3(3n+1)}

\frac{n}{3n+1}

Therefore, sum of the given series up to n terms is \frac{n}{3n+1} .

Question 10. \frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.14}+\frac{1}{14.19}....\frac{1}{(5n-4)(5n+1)}  to n terms.

Solution: 

We are given the nth term of the given series,

an\frac{1}{(5n-4)(5n+1)}

\frac{1}{5}\left[\frac{1}{5n-4}-\frac{1}{5n+1}\right]

Now we have to summate our nth term to find the sum(Sn) of this series.

S_n = \sum_{k=1}^{n} a_k = \frac{1}{5}\sum_{k=1}^{n} \left[\frac{1}{5n-4}-\frac{1}{5n+1}\right]

\frac{1}{3}\left[(1-\frac{1}{6})+(\frac{1}{6}-\frac{1}{11})+(\frac{1}{11}-\frac{1}{14})...(\frac{1}{5n-4}-\frac{1}{5n+1})\right]

\frac{1}{5}\left[1-\frac{1}{5n+1}\right]

\frac{5n}{5(5n+1)}

\frac{n}{5n+1}

Therefore, sum of the given series up to n terms is \frac{n}{5n+1} .



Last Updated : 30 Apr, 2021
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