The Dance of the Series
Welcome, fellow traveler on the JEE journey. Today, we are not just solving a problem; we are uncovering the hidden rhythm of a series.
Look at the expression: 2⋅22−32+2⋅42−52+2⋅62−…. At first glance, it looks like a chaotic jumble of squares and alternating signs.
In the world of JEE, chaos is just a pattern waiting to be discovered. Imagine you are standing before this series, and you see the terms dancing in pairs.
The first term, 2⋅22, is positive and even-based. The second term, −32, is negative and odd-based. This pattern repeats perfectly.
Since we have twenty terms in total, we can group them into exactly ten pairs. This is the first step of our masterclass: the art of grouping.
The Algebraic Transformation
Now, let us define the general term for these pairs, which we will call Tr. For the r-th pair, the even term is 2⋅(2r)2 and the odd term is −(2r+1)2.
So, our general pair is Tr=2⋅(2r)2−(2r+1)2. Let's expand this carefully.
2⋅(2r)2 becomes 2⋅4r2, which is 8r2. The second part, (2r+1)2, expands to 4r2+4r+1.
When we subtract this, we get Tr=8r2−(4r2+4r+1). Simplifying this, we arrive at the elegant expression:
This is the soul of our series, the engine that drives the entire calculation.
The Power of Summation
Now that we have our simplified general term, we need to sum these ten pairs. We apply the summation operator:
Using the linearity of summation, we can split this into three distinct parts:
S=4r=1∑10r2−4r=1∑10r−r=1∑101
This is where our toolkit comes in. We recall the standard formulas:
r=1∑nr2=6n(n+1)(2n+1)andr=1∑nr=2n(n+1)
With n=10, these become:
610⋅11⋅21=385and210⋅11=55
The Final Calculation
We are almost there. Substituting these values back into our equation, we get:
Calculating these, we find 4⋅385=1540 and 4⋅55=220.
Finally, we compute the total:
There it is! The chaos has been tamed, and the answer is 1310. Remember, every complex problem is just a series of simple steps waiting for you to take them.