Question

7. Find the degree of the expression (1+x)(1+x^6)(1+x^11)....(1+x^101)
A) 1070
C) 1072
B) 1071
D) 1073​

Answer 1

Hey there!

Given,

$(1 + x)(1 + x^6)(1+x^{11}).......(1+x^{101})$

This expression goes like;

$\displaystyle \prod_{i=0}^{20} (1 + x^{1 + 5i})$

Thedegree of expression will be equal to the sum of the powers of x in all the factors. i.e 1 + 6 + 11 .......101.

It forms an Arithmetic Progression with first term 1 and common difference = 6-1 = 11-6 = 5.

Since, last term is 101. We can find number of terms.

$T_n = 101\\\\a+(n-1)d = 101\\\\1 + (n-1)5 = 101\\\\1 + 5n - 5 = 101\\\\5n = 101 + 4\\\\n = 21$

Now, general formula of Sn;

$S_n = \dfrac{n}{2}[2a + (n-1)d]\\\\S_{21} = \dfrac{21}{2}[2 \times 1 + 20 \times 5]\\\\S_{21} = \dfrac{21}{2} \times 102\\\\S_{21} = 21 \times 51\\\\\boxed{S_{21} = 1071}$

Hence, degree of expression is 1071.

Answer 2

Answer:

We want to determine the degree of the expression (1+x)(1+x6)(1+x11)⋯(1+x100).(1+x)(1+x6)(1+x11)⋯(1+x100).

There is an error in the question.

It can be seen that the first three factors are of the type (1+x1+5k)(1+x1+5k). This is followed by three dots indicating that the other factors are also of the same type.

However the last factor, (1+x100),(1+x100), is not of this type since there is no integer, k,k, satisfying the requirement 1+5k=100.1+5k=100.

I am presuming that this is a typographical error and that the last factor should be (1+x101).(1+x101).

So, the expression is ∏k=020(1+x1+5k).∏k=020(1+x1+5k).

The highest power of xx in this expression is the sum of the powers of xx in all the factors

=1+6+11+⋯+101.=1+6+11+⋯+101.

This is an Arithmetic Progression having the first term a=1a=1 and common difference d=5.d=5. This AP has 2121 terms since the last term, n,n, is 1+20×5=101.1+20×5=101.

The sum of an AP is Sn=n2(t1+tn)Sn=n2(t1+tn)