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 + an . . . .(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 + an . . . .(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} .
