Mathematics Test - 26

The given series is a G.P with a common ratio \(\operatorname{of}(1+i)\)
Hence \(1+(1+i)+(1+i)^{2}+(1+i)^{3}\)
\(=\frac{1-(1+i)^{4}}{1-(1+i)}\)
Now \(1+i=\sqrt{2}\left(\frac{1+i}{\sqrt{2}}\right)\)
\(=\sqrt{2}\left(e^{i \frac{\pi}{4}}\right)\)
Hence \(\frac{1-(1+i)^{4}}{-i}\)
\(=\frac{1-\left(\sqrt{2}\left(e^{i \frac{\pi}{4}}\right)\right)^{4}}{-i}\)
\(=\frac{1-4\left(e^{i \pi}\right)}{-i}\)
\(=\frac{1-4(-1)}{-i}\)
\(=\frac{5}{-i}\)
\(=5 i\)
Hence, the correct option is (B)

\(S=i^{2}+i^{4}+i^{6}+\ldots i^{2(2 n+1)}\)
The above is a G.p series with a common ratio
of \(i^{2}\)
Hence
\(S_{2 n+1}=\frac{i^{2}\left(1-i^{2(2 n+1)}\right)}{1-i^{2}}\)
\(=\frac{i^{2}\left(1-i^{4 n+2}\right)}{1-(-1)}\)
\(=\frac{i^{2}\left(1-i^{2} \cdot i^{4 n}\right)}{2}\)
\(=\frac{i^{2}\left(1+i^{4 n}\right)}{2}\)
\(=\frac{-\left(1+\left(i^{4}\right)^{n}\right)}{2}\)
\(=\frac{-(1+1)}{2}\)
\(=-1\)
Hence, the correct option is (A)

From \(p m\) 's, we can have any of \(0,1,2, \ldots p m\) 's. From \(p l\) s we can have any of \(0,1, \ldots, p l\) s. Besides these for remaining \(p\) letters we have only two combinations namely 0 or \(1 .\) So the required number of combinations is \(=\) Coefficient of \(x^{p}\) in \(\left(1+x+x^{2}+x^{3}+\ldots+x^{p}\right)\left(1+x+x^{2}+\ldots+x^{p}\right) \frac{(1+x)(1+x) \ldots \ldots(1+x)}{p-\text { factors }}\)
\(=\) Coefficient of \(x^{p}\) in \(\frac{\left(1-x^{p+1}\right)^{2}}{(1-x)^{2}}(1+x)^{p}\)
\(=\) Coefficient of \(x^{p}\) in \(\left(1-x^{p+1}\right)^{2}(1-x)^{-2}(1+x)^{p}\)
\(=\) coefficient of \(x^{p}\) in \((1+x)^{p}(1-x)^{-2}\)
As \(\left(1-x^{p+1}\right)^{2}\) does not contain \(x^{p}\) =Coefficient of \(x^{p}\) in \([2-(1-x)]^{p}(1-x)^{-2}\) \(=\) coefficient of \(x^{p}\) in \(2 p(1-x)^{-2}-p 2^{p-1}(1-x)^{-1}+^{p} C_{2} 2^{r-2}(1-x)^{0} \rightarrow C_{3} 2^{p-3}(1-x)+\ldots+(-1)^{p} \cdot P C_{p}(1-x)^{p-2}\)
=coefficient of \(x^{p}\) in \(2^{p}(1-x)^{-2}-p \cdot 2^{p-1}(1-x)^{-1}\) \(=2^{p}+p 2^{p-1}\)
Hence, the correct option is (D)

Given, \(x+y+z+u+t=20 \ldots \ldots \ldots . .(1)\)
and \(x+y+z=5\)
The given system of equations can be written
as
\(u+t=15 \ldots \ldots . .(3)\)
\(x+y+z=5 \ldots \ldots \ldots \ldots .(4)\)
No. of non-negative integral solutions of (4) are \(n+r-1 c_{r}=3+5-1 c_{5}=7 c_{3}\)
No. of non-negative integral solutions of ( 3 ) are \({ }^{n+r-1} c_{r}={ }^{2+15-1} c_{15}={ }^{16} c_{15}\)
So, required numbers \(={ }^{16} c_{15} \times{ }^{7} c_{5}=336\)
Hence, the correct option is (A)

There are two possible cases Case 1 : Five 1 's, one 2 's, one 3 's Number of numbers \(=\frac{7 !}{5 !}=42\) Case 2 : Four 1 's, three 2 's Number of numbers \(=\frac{7 !}{4 ! 3 !}=35\)
Total number of numbers \(=42+35=77\)
Hence, the correct option is (C)

It is a circular permutation,
So, \((5\) of them are identical)
\(=\frac{8 !}{2.5 !}\)
\(=\frac{8.7 .6}{2}\)
\(=4.7 .6\)
\(=168\)
Hence, the correct option is (D)

64 can be written as \(64^{1}, 8^{2}, 4^{3}, 2^{6}\)
As \(x>y\) Possible options of writing 64 are \(64^{1}, 8^{2}, 4^{3}\)
since, \(x \neq 4\) From the chosen options, Possible options of writing 64 are \(64^{1}, 8^{2}\)
since, \(y \neq 1\) From the chosen options, Possible options of writing 64 is only \(8^{2}\)
So, \(x=8, y=2\)
\(=>x+y=10\)
Hence, the correct option is (D)

\((p+1),(p+2),(p+3), \ldots .(p+q)\) are \(q\) consecutive numbers. \((p+1)(p+2)(p+3) \ldots(p+q)\) is product of \(^{\prime} q^{\prime}\) consecutive natural numbers. The product will be divisible by all its factors and since it is a product of consecutive natural numbers, so it will always be divisible by \(q !\) Also, \(q !\) is the greatest integer amongst all their divisors. Hence, the greatest integer which divides the product is \(q !\)
Hence, the correct option is (B)

Highest power of 3 contained in \(120 !\) \(=\left[\frac{120}{3}\right]+\left[\frac{120}{3^{2}}\right]+\left[\frac{120}{3^{3}}\right]+\left[\frac{120}{3^{4}}\right]+\left[\frac{120}{3^{5}}\right]\)
\(=40+13+4+1+0\)
\(=58\)
Hence, the correct option is (D)