We define the natural logarithm of a postive number \(a\) as \(\log_e a = \ln a\). Logarithm of a negative number is not defined.
Let’s recollect the basic properties of natural logarithms:
- \(\ln ab = \ln a+ \ln b\)
- \(\ln \frac{a}{b} = \ln a – \ln b\)
- \(\ln a^x=x\ln a\)
- \(\ln ab = \ln a+\ln b\)
- \(\ln \frac{a}{b} = \ln a – \ln b\)
- \(\log _b a = \log _ca. \log _bc\)
- \(a^{\log _ax}=x\)
- \(\log _{a^k}(x)=\frac{1}{k}\log_a (x)\)
- \(\ln 1=0\)
Ex -1
If \(a>1,b>1,c>1\) be 3 reals in GP, show that \(\frac{1}{1+\ln a}, \frac{1}{1+ \ln b}, \frac{1}{1+ \ln c}\) are in HP
Solution:
We have \(b^2=ac\thinspace \Rightarrow \ln b = \frac{\ln a+\ln b}{2}\), which implies \(\ln a, \ln b, \ln c\) are in AP. This also confirms that \(1+\ln a, 1+\ln b, 1+\ln c\) are in AP, and the conclusion that the required terms are in HP follows from this.
2.1 Sum of special sequences
Consider the sequence \(a_n\) defined by \(a_i=i^2\). How do we find the sum of n terms of this sequence?
What about the sum of the sequence \(a_n\) defined by \(a_i=i\)? This seuence is in AP, and the sum of the first \(n\) terms = \(\frac{n(n+1)}{2}\), which should be obvious by now from what we have studied in our earlier module.
To find the sum of the squares of the first \(n\) natural numbers, we observe the following:
\(n^3-(n-1)^3=3n^2-3n+1\). If we sum upto n terms either sides of the equality sign, we have
\(n^3=3(1^2+2^2+….+n^2)-3(1+2+3+….+n)+n\). Let’s call \(1^2+2^2+…+n^2=S\), so
\(3S=n^3+\frac{3n(n+1)}{2}-n\thinspace \Rightarrow S=\frac{n(n+1)(2n+1)}{6}\).
Let’s find the sum of cubes of the first n natural numbers in the same way:
We observe that \(n^4-(n-1)^4=4n^3-6n^2+4n-1\). Again summing upto n terms on either sides, we have \(n^4=4(1^3+2^3+…+n^3)-6(1^2+2^2+…+n^2)+4(1+2+3…+n)-n\). Calling \(1^3+2^3+…+n^3=S\), we have \(4S=n^4+n+6(1^2+2^2+…+n^2)-4(1+2+…+n)\). Putting in the expression for the sum of squares and sum of first n positive integers we have \(S=\left \{ \frac{n(n+1)}{2} \right \}^2\).
Ex -2
Find the sum of the series : \(\frac{1^2}{1}+\frac{1^2+2^2}{1+2}+\frac{1^2+2^2+3^2}{1+2+3}+…+\frac{1^2+2^2+…+n^2}{1+2+…+n}\).
Solution:
General term of this sequence : \(T_r=\frac{1^2+2^2+…+r^2}{1+2+…r}=\frac{2r(r+1)(2r+1)}{6r(r+1)}=\frac{1}{3}(2r+1)\). Thus we have
\(\sum\limits_{r=1}^{n}T_r=\frac{2}{3}\left(\sum\limits_{r=1}^{n}r \right )+\frac{n}{3}\).
We thus have the sum as \(\frac{n(n+1)}{3}+\frac{n}{3}=\frac{n(n+2)}{3}\).
Ex-3
Find the sum upto n terms of the series \(1.2.3+2.3.4+3.4.5+…\)
Solution:
General term of this sequence : \(T_k=k(k+1)(k+2)=k^3+3k^2+2k\).
Thus \(\sum\limits_{k=1}^nT_k=\sum\limits_{k=1}^nk^3+3\sum\limits_{k=1}^nk^2+2\sum\limits_{k=1}^nk\). Plugging in the expressions for the sum of cubes, squares and first n natural numbers we have the sum as
Upon simplification this turns out to