0I will solve a few questions which most students do in the wrong way.
Firstly, I will tell what students do and then the correct solution
1)The sum to infinity of the series
1+2/3+6/3^2+10/3^3+...
Here mostly student will apply AGP formula taking a=2,r=1/3, and d=4.
And get the result as seven, which is wrong.
But what's the mistake,
Many students only concentrate on the formula they sometimes forget to look from where the formula has arrived.
So the mistake is in taking 'a' as 2
As to use AGP formula directly the condition is the series should be like a,(a+d).r, (a+2d).r^2,...
And in given AGP terms are like 1,a.r, (a+d).r^2,...
So to use the formula, we have to alter it so that the conditions satisfy
Taking 2/3 common from the series, we get a=1,d=4,r=1/3 and after solving we get the correct answer.
2) Let a,b,c,d be four positive real numbers such that their product is unity. Then the least value of (1+a)(1+b)(1+c)(1+d)
Mostly student use the trick as a,b,c,d >0 so lets put a=b=c=d=1 and find the answer.
In this case, it holds true, but it will not always, so we should use the correct logic. We all know that A.M>G.M, so using that here
(1+a)/2>1.a =a which implies (1+a)>2a similarly (1+b)>2b, (1+c)>2c, (1+d)>2d.
Which implies
(1+a)(1+b)(1+c)(1+d)>2a.2b.2c.2d
=2.2.2.2.abcd
And given abcd=1
Which implies
(1+a)(1+b)(1+c)(1+d)>2.2.2.2=16
Hence 16 is the least value.
Like these, students make many mistakes, and in my classes, I'll try to resolve all the problems/errors.
I hope you enjoyed it.
