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Question: The sum of \(50\) terms of the series \(1 + 2\left( {1 + \dfrac{1}{{50}}} \right) + 3{\left( {1 + \d...

The sum of 5050 terms of the series 1+2(1+150)+3(1+150)2+4(1+150)3+....1 + 2\left( {1 + \dfrac{1}{{50}}} \right) + 3{\left( {1 + \dfrac{1}{{50}}} \right)^2} + 4{\left( {1 + \dfrac{1}{{50}}} \right)^3} + .... is given by
A) 25502550
B) 25002500
C) 24502450
D) None of these

Explanation

Solution

In the question, we are required to find the sum of an arithmetic geometric progression. So, we follow a structured method to find the same. Arithmetic geometric progression is a series whose terms involve the product of two types of terms. One term is in arithmetic progression and the other one is in geometric progression.

Complete step by step solution:
We need to find the sum of the given arithmetic geometric progression up to fifty terms.
So, S=1+2(1+150)+3(1+150)2+4(1+150)3+....50(1+150)49S = 1 + 2\left( {1 + \dfrac{1}{{50}}} \right) + 3{\left( {1 + \dfrac{1}{{50}}} \right)^2} + 4{\left( {1 + \dfrac{1}{{50}}} \right)^3} + ....50{\left( {1 + \dfrac{1}{{50}}} \right)^{49}}
In this arithmetic geometric progression, the (1+150)\left( {1 + \dfrac{1}{{50}}} \right) component is in geometric progression and constant term is arithmetic progression.
Hence, we multiply both sides of the equation by common ratio of the geometric progressive terms, (1+150)\left( {1 + \dfrac{1}{{50}}} \right).
So, S=1+2(1+150)+3(1+150)2+4(1+150)3+....50(1+150)49−−−−−−−−(1)S = 1 + 2\left( {1 + \dfrac{1}{{50}}} \right) + 3{\left( {1 + \dfrac{1}{{50}}} \right)^2} + 4{\left( {1 + \dfrac{1}{{50}}} \right)^3} + ....50{\left( {1 + \dfrac{1}{{50}}} \right)^{49}} - - - - - - - - (1)
⇒(1+150)S=(1+150)[1+2(1+150)+3(1+150)2+4(1+150)3+....50(1+150)49]\Rightarrow \left( {1 + \dfrac{1}{{50}}} \right)S = \left( {1 + \dfrac{1}{{50}}} \right)\left[ {1 + 2\left( {1 + \dfrac{1}{{50}}} \right) + 3{{\left( {1 + \dfrac{1}{{50}}} \right)}^2} + 4{{\left( {1 + \dfrac{1}{{50}}} \right)}^3} + ....50{{\left( {1 + \dfrac{1}{{50}}} \right)}^{49}}} \right]
Opening the bracket and simplifying, we get,
⇒(1+150)S=(1+150)+2(1+150)2+3(1+150)3+4(1+150)4+....50(1+150)50\Rightarrow \left( {1 + \dfrac{1}{{50}}} \right)S = \left( {1 + \dfrac{1}{{50}}} \right) + 2{\left( {1 + \dfrac{1}{{50}}} \right)^2} + 3{\left( {1 + \dfrac{1}{{50}}} \right)^3} + 4{\left( {1 + \dfrac{1}{{50}}} \right)^4} + ....50{\left( {1 + \dfrac{1}{{50}}} \right)^{50}}
Now, subtracting (2)(2) from (1)(1), we get,
\Rightarrow S - \left( {1 + \dfrac{1}{{50}}} \right)S = \left[ {1 + 2\left( {1 + \dfrac{1}{{50}}} \right) + 3{{\left( {1 + \dfrac{1}{{50}}} \right)}^2} + 4{{\left( {1 + \dfrac{1}{{50}}} \right)}^3} + ....50{{\left( {1 + \dfrac{1}{{50}}} \right)}^{49}}} \right] - \left[ {\left( {1 + \dfrac{1}{{50}}} \right) + 2{{\left( {1 + \dfrac{1}{{50}}} \right)}^2} + 3{{\left( {1 + \dfrac{1}{{50}}} \right)}^3} + 4{{\left( {1 + \dfrac{1}{{50}}} \right)}^4} + ....50{{\left( {1 + \dfrac{1}{{50}}} \right)}^{50}}} \right]$$$$ \Rightarrow S - S - \dfrac{S}{{50}} = \left[ {1 + \left( {1 + \dfrac{1}{{50}}} \right) + {{\left( {1 + \dfrac{1}{{50}}} \right)}^2} + {{\left( {1 + \dfrac{1}{{50}}} \right)}^3} + ....{{\left( {1 + \dfrac{1}{{50}}} \right)}^{49}}} \right] - 50{\left( {1 + \dfrac{1}{{50}}} \right)^{50}}
The term contained in square brackets is a geometric progression. Hence, we can find the sum of n terms of a geometric progression by formula: a(rn−1)(r−1)a\dfrac{{\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}
⇒−S50=1((1+150)50−1)((1+150)−1)−50(1+150)50\Rightarrow - \dfrac{S}{{50}} = 1\dfrac{{\left( {{{\left( {1 + \dfrac{1}{{50}}} \right)}^{50}} - 1} \right)}}{{\left( {\left( {1 + \dfrac{1}{{50}}} \right) - 1} \right)}} - 50{\left( {1 + \dfrac{1}{{50}}} \right)^{50}}
Simplifying further, we get,
⇒−S50=((1+150)50−1)(1+150−1)−50(1+150)50\Rightarrow - \dfrac{S}{{50}} = \dfrac{{\left( {{{\left( {1 + \dfrac{1}{{50}}} \right)}^{50}} - 1} \right)}}{{\left( {1 + \dfrac{1}{{50}} - 1} \right)}} - 50{\left( {1 + \dfrac{1}{{50}}} \right)^{50}}
⇒−S50=50((1+150)50−1)−50(1+150)50\Rightarrow - \dfrac{S}{{50}} = 50\left( {{{\left( {1 + \dfrac{1}{{50}}} \right)}^{50}} - 1} \right) - 50{\left( {1 + \dfrac{1}{{50}}} \right)^{50}}
⇒−S50=50(1+150)50−50−50(1+150)50\Rightarrow - \dfrac{S}{{50}} = 50{\left( {1 + \dfrac{1}{{50}}} \right)^{50}} - 50 - 50{\left( {1 + \dfrac{1}{{50}}} \right)^{50}}
Cancelling terms having same magnitude and opposite signs, we get,
⇒−S50=−50\Rightarrow - \dfrac{S}{{50}} = - 50
Cross multiplying the terms to find the value of S, we get,
⇒S=2500\Rightarrow S = 2500
Therefore, the sum of fifty terms of the series 1+2(1+150)+3(1+150)2+4(1+150)3+....1 + 2\left( {1 + \dfrac{1}{{50}}} \right) + 3{\left( {1 + \dfrac{1}{{50}}} \right)^2} + 4{\left( {1 + \dfrac{1}{{50}}} \right)^3} + .... is 25002500. Hence, option (B) is the correct answer.

Note:
The method involved in the problem given to us must be remembered as it is a standard method that can be applied in various questions involving arithmetic geometric progressions or any other type of special series or progression. We also must remember the formula for finding the sum of n terms of a geometric progression: a(1−rn)(1−r)a\dfrac{{\left( {1 - {r^n}} \right)}}{{\left( {1 - r} \right)}} and the formula for finding the sum of an infinite geometric progression a1−r\dfrac{a}{{1 - r}}.