The difference between C.I. and S.I. on a certain sum of money at 10% per annum for 3 years is Rs.620. find the principle if it is known that the interest is compounded annually.

Hint: Here in this problem we are given with difference in C.I. and S.I. and the rate of interest with period. Now we are going to use the formulas directly in order to find the principal value.

Formulas used:
1. Simple interest = \[\dfrac{{PRT}}{{100}}\]
2. Compound interest = \[P\left( {{{\left( {1 + \dfrac{R}{{100}}} \right)}^T} - 1} \right)\]

Step by step solution:
Given that,
Time period T=3years
Rate R=10%
Now the difference in C.I. and S.I. is given by
\[ \Rightarrow P\left( {{{\left( {1 + \dfrac{R}{{100}}} \right)}^T} - 1} \right) - \dfrac{{PRT}}{{100}}\]
Putting the values of T and R
\[ \Rightarrow P\left( {{{\left( {1 + \dfrac{{10}}{{100}}} \right)}^3} - 1} \right) - \dfrac{{P \times 10 \times 3}}{{100}}\]
Using the algebraic identity \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}\] we get
\[ \Rightarrow P\left( {1 + {{\left( {\dfrac{{10}}{{100}}} \right)}^3} + 3 \times \dfrac{{10}}{{100}} + 3 \times {{\left( {\dfrac{{10}}{{100}}} \right)}^2} - 1} \right) - \dfrac{{P \times 10 \times 3}}{{100}}\]
Cancelling 1 from above expansion
\[ \Rightarrow P\left( {{{\left( {\dfrac{{10}}{{100}}} \right)}^3} + \dfrac{{30}}{{100}} + 3 \times {{\left( {\dfrac{{10}}{{100}}} \right)}^2}} \right) - \dfrac{{P \times 10 \times 3}}{{100}}\]
Multiplying with P,
\[ \Rightarrow P{\left( {\dfrac{{10}}{{100}}} \right)^3} + \dfrac{{30P}}{{100}} + 3P{\left( {\dfrac{{10}}{{100}}} \right)^2} - \dfrac{{30P}}{{100}}\]
Cancelling second and last term we get,
\[ \Rightarrow P{\left( {\dfrac{{10}}{{100}}} \right)^3} + 3P{\left( {\dfrac{{10}}{{100}}} \right)^2}\]
\[ \Rightarrow P{\left( {\dfrac{{10}}{{100}}} \right)^2}\left( {\dfrac{1}{{10}} + 3} \right)\]
Taking LCM of second bracket
\[ \Rightarrow P \times \dfrac{1}{{100}} \times \dfrac{{31}}{{10}}\]
\[ \Rightarrow P \times \dfrac{{31}}{{1000}}\]
But this difference is already given to us as Rs. 620
\[ \Rightarrow P \times \dfrac{{31}}{{1000}} = 620\]
\[ \Rightarrow P = 620 \times \dfrac{{1000}}{{31}}\]
Cancelling by 31 we get
\[ \Rightarrow P = 20 \times 1000\]
\[ \Rightarrow P = 20000\]
This is the principle amount \[ \Rightarrow P = 20000\]

Note:
In this problem instead of using or expanding the bracket we can directly use the formula if 3 years of compound interest are given \[\dfrac{{{{\Pr }^2}}}{{{{100}^3}}}\left( {300 + r} \right)\]. Otherwise we have solved it with traditional methods.