Question

The coefficient of ${{t}^{50}}$ in ${{\left( 1+{{t}^{2}} \right)}^{25}}\left( 1+{{t}^{25}} \right)\left( 1+{{t}^{40}} \right)\left( 1+{{t}^{45}} \right)\left( 1+{{t}^{47}} \right)$ is
(A) $1{{+}^{25}}{{C}_{5}}$
(B) $1{{+}^{25}}{{C}_{5}}{{+}^{25}}{{C}_{7}}$
(C) $1{{+}^{25}}{{C}_{7}}$
(D) none of these

Answer 1

First of all, expand the term ${{\left( 1+{{t}^{2}} \right)}^{25}}$ using binomial expansion formula, ${{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( 1 \right)}^{n}}{{\left( x \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( 1 \right)}^{n-1}}{{\left( x \right)}^{1}}{{+}^{n}}{{C}_{2}}{{\left( 1 \right)}^{n-2}}{{\left( x \right)}^{2}}{{+}^{n}}{{C}_{3}}{{\left( 1 \right)}^{n-3}}{{\left( x \right)}^{3}}{{+}^{n}}{{C}_{4}}{{\left( 1 \right)}^{n-4}}{{\left( x \right)}^{4}}+$
$^{n}{{C}_{5}}{{\left( 1 \right)}^{n-5}}{{\left( x \right)}^{5}}+...................{{+}^{n}}{{C}_{n}}{{\left( 1 \right)}^{0}}{{\left( x \right)}^{n}}$ . Now, transform the expression ${{\left( 1+{{t}^{2}} \right)}^{25}}\left( 1+{{t}^{25}} \right)\left( 1+{{t}^{40}} \right)\left( 1+{{t}^{45}} \right)\left( 1+{{t}^{47}} \right)$ as,
\left\\{ ^{25}{{C}_{0}}{{+}^{25}}{{C}_{1}}\left( {{t}^{2}} \right){{+}^{25}}{{C}_{2}}\left( {{t}^{4}} \right){{+}^{25}}{{C}_{3}}\left( {{t}^{6}} \right){{+}^{25}}{{C}_{4}}\left( {{t}^{8}} \right){{+}^{25}}{{C}_{5}}\left( {{t}^{10}} \right)+...............{{+}^{25}}{{C}_{25}}\left( {{t}^{50}} \right) \right\\}\left( 1+{{t}^{25}} \right)
$\times \left( 1+{{t}^{40}} \right)\left( 1+{{t}^{45}} \right)\left( 1+{{t}^{47}} \right)$ . The term ${{t}^{50}}$can be obtained by expanding the term ${{\left( 1+{{t}^{2}} \right)}^{25}}$ and also by multiplying the term ${{t}^{40}}$ of $\left( 1+{{t}^{40}} \right)$ with the term $^{25}{{C}_{5}}\left( {{t}^{10}} \right)$ of the term \left\\{ ^{25}{{C}_{0}}{{+}^{25}}{{C}_{1}}\left( {{t}^{2}} \right){{+}^{25}}{{C}_{2}}\left( {{t}^{4}} \right){{+}^{25}}{{C}_{3}}\left( {{t}^{6}} \right){{+}^{25}}{{C}_{4}}\left( {{t}^{8}} \right){{+}^{25}}{{C}_{5}}\left( {{t}^{10}} \right)+...............{{+}^{25}}{{C}_{25}}\left( {{t}^{50}} \right) \right\\} . Now, solve it further and get the coefficient of ${{t}^{50}}$ .

Complete step by step answer:
According to the question, we have the expression,
${{\left( 1+{{t}^{2}} \right)}^{25}}\left( 1+{{t}^{25}} \right)\left( 1+{{t}^{40}} \right)\left( 1+{{t}^{45}} \right)\left( 1+{{t}^{47}} \right)$ …………………………………(1)
We have to get the coefficient of ${{t}^{50}}$ in the above expression.
From equation (1), there is an expression that has the term ${{\left( 1+{{t}^{2}} \right)}^{25}}$ . We need to expand the term ${{\left( 1+{{t}^{2}} \right)}^{25}}$ .
We know the formula for the binomial expansion,
${{\left( 1+x \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( 1 \right)}^{n}}{{\left( x \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( 1 \right)}^{n-1}}{{\left( x \right)}^{1}}{{+}^{n}}{{C}_{2}}{{\left( 1 \right)}^{n-2}}{{\left( x \right)}^{2}}{{+}^{n}}{{C}_{3}}{{\left( 1 \right)}^{n-3}}{{\left( x \right)}^{3}}{{+}^{n}}{{C}_{4}}{{\left( 1 \right)}^{n-4}}{{\left( x \right)}^{4}}$
${{+}^{n}}{{C}_{5}}{{\left( 1 \right)}^{n-5}}{{\left( x \right)}^{5}}+...................{{+}^{n}}{{C}_{n}}{{\left( 1 \right)}^{0}}{{\left( x \right)}^{n}}$ ………………………………………(6)
Using the binomial expansion formula shown in equation (6), expanding the term ${{\left( 1+{{t}^{2}} \right)}^{25}}$ , we get
${{\left( 1+{{t}^{2}} \right)}^{25}}{{=}^{25}}{{C}_{0}}{{\left( 1 \right)}^{25}}{{\left( {{t}^{2}} \right)}^{0}}{{+}^{25}}{{C}_{1}}{{\left( 1 \right)}^{25-1}}{{\left( {{t}^{2}} \right)}^{1}}{{+}^{25}}{{C}_{2}}{{\left( 1 \right)}^{25-2}}{{\left( {{t}^{2}} \right)}^{2}}{{+}^{25}}{{C}_{3}}{{\left( 1 \right)}^{25-3}}{{\left( {{t}^{2}} \right)}^{3}}{{+}^{25}}{{C}_{4}}{{\left( 1 \right)}^{25-4}}{{\left( {{t}^{2}} \right)}^{4}}+$
$^{25}{{C}_{5}}{{\left( 1 \right)}^{25-5}}{{\left( {{t}^{2}} \right)}^{5}}+...............{{+}^{25}}{{C}_{25}}{{\left( 1 \right)}^{0}}{{\left( {{t}^{2}} \right)}^{25}}$
$\Rightarrow {{\left( 1+{{t}^{2}} \right)}^{25}}{{=}^{25}}{{C}_{0}}\left( 1 \right){{\left( {{t}^{2}} \right)}^{0}}{{+}^{25}}{{C}_{1}}\left( 1 \right){{\left( {{t}^{2}} \right)}^{1}}{{+}^{25}}{{C}_{2}}\left( 1 \right){{\left( {{t}^{2}} \right)}^{2}}{{+}^{25}}{{C}_{3}}\left( 1 \right){{\left( {{t}^{2}} \right)}^{3}}{{+}^{25}}{{C}_{4}}\left( 1 \right){{\left( {{t}^{2}} \right)}^{4}}+$
$^{25}{{C}_{5}}\left( 1 \right){{\left( {{t}^{2}} \right)}^{5}}+...............{{+}^{25}}{{C}_{25}}{{\left( 1 \right)}^{0}}{{\left( {{t}^{2}} \right)}^{25}}$
$\Rightarrow {{\left( 1+{{t}^{2}} \right)}^{25}}{{=}^{25}}{{C}_{0}}{{+}^{25}}{{C}_{1}}\left( {{t}^{2}} \right){{+}^{25}}{{C}_{2}}\left( {{t}^{4}} \right){{+}^{25}}{{C}_{3}}\left( {{t}^{6}} \right){{+}^{25}}{{C}_{4}}\left( {{t}^{8}} \right){{+}^{25}}{{C}_{5}}\left( {{t}^{10}} \right)+...............{{+}^{25}}{{C}_{25}}\left( {{t}^{50}} \right)$ ………………………………..(7)
Now, putting the value of ${{\left( 1+{{t}^{2}} \right)}^{25}}$ from equation (7) in equation (1), we get
\left\\{ ^{25}{{C}_{0}}{{+}^{25}}{{C}_{1}}\left( {{t}^{2}} \right){{+}^{25}}{{C}_{2}}\left( {{t}^{4}} \right){{+}^{25}}{{C}_{3}}\left( {{t}^{6}} \right){{+}^{25}}{{C}_{4}}\left( {{t}^{8}} \right){{+}^{25}}{{C}_{5}}\left( {{t}^{10}} \right)+...............{{+}^{25}}{{C}_{25}}\left( {{t}^{50}} \right) \right\\}\left( 1+{{t}^{25}} \right)$$$$\times \left( 1+{{t}^{40}} \right)\left( 1+{{t}^{45}} \right)\left( 1+{{t}^{47}} \right) ………………………………..(8)
We have to get the coefficient of ${{t}^{50}}$ .
First of all, we need to find ways to obtain ${{t}^{50}}$ from the expression,
\left\\{ ^{25}{{C}_{0}}{{+}^{25}}{{C}_{1}}\left( {{t}^{2}} \right){{+}^{25}}{{C}_{2}}\left( {{t}^{4}} \right){{+}^{25}}{{C}_{3}}\left( {{t}^{6}} \right){{+}^{25}}{{C}_{4}}\left( {{t}^{8}} \right){{+}^{25}}{{C}_{5}}\left( {{t}^{10}} \right)+...............{{+}^{25}}{{C}_{25}}\left( {{t}^{50}} \right) \right\\}\left( 1+{{t}^{25}} \right)
$\times \left( 1+{{t}^{40}} \right)\left( 1+{{t}^{45}} \right)\left( 1+{{t}^{47}} \right)$
In the above equation, we have ${{t}^{50}}$ in the term \left\\{ ^{25}{{C}_{0}}{{+}^{25}}{{C}_{1}}\left( {{t}^{2}} \right){{+}^{25}}{{C}_{2}}\left( {{t}^{4}} \right){{+}^{25}}{{C}_{3}}\left( {{t}^{6}} \right){{+}^{25}}{{C}_{4}}\left( {{t}^{8}} \right){{+}^{25}}{{C}_{5}}\left( {{t}^{10}} \right)+...............{{+}^{25}}{{C}_{25}}\left( {{t}^{50}} \right) \right\\} .
If we multiply the term ${{t}^{50}}$ with the terms ${{t}^{25}},{{t}^{40}},{{t}^{45}}$ , and ${{t}^{47}}$ , we will get the exponent of t greater than 50. Therefore, to get the term ${{t}^{50}}$ we have to multiply the term $^{25}{{C}_{25}}\left( {{t}^{50}} \right)$ only with number 1 that is present in the term $\left( 1+{{t}^{25}} \right)$ , $\left( 1+{{t}^{40}} \right)$ , $\left( 1+{{t}^{45}} \right)$ , and $\left( 1+{{t}^{47}} \right)$ of the expression, \left\\{ ^{25}{{C}_{0}}{{+}^{25}}{{C}_{1}}\left( {{t}^{2}} \right){{+}^{25}}{{C}_{2}}\left( {{t}^{4}} \right){{+}^{25}}{{C}_{3}}\left( {{t}^{6}} \right){{+}^{25}}{{C}_{4}}\left( {{t}^{8}} \right){{+}^{25}}{{C}_{5}}\left( {{t}^{10}} \right)+...............{{+}^{25}}{{C}_{25}}\left( {{t}^{50}} \right) \right\\}\left( 1+{{t}^{25}} \right)
$\times \left( 1+{{t}^{40}} \right)\left( 1+{{t}^{45}} \right)\left( 1+{{t}^{47}} \right)$ and ignore the remaining terms.
In this case, the coefficient of ${{t}^{50}}$ is $^{25}{{C}_{25}}$ ……………………………(9)
The term ${{t}^{50}}$ can also be obtained by multiplying the term ${{t}^{40}}$ of $\left( 1+{{t}^{40}} \right)$ with the term $^{25}{{C}_{5}}\left( {{t}^{10}} \right)$ of \left\\{ ^{25}{{C}_{0}}{{+}^{25}}{{C}_{1}}\left( {{t}^{2}} \right){{+}^{25}}{{C}_{2}}\left( {{t}^{4}} \right){{+}^{25}}{{C}_{3}}\left( {{t}^{6}} \right){{+}^{25}}{{C}_{4}}\left( {{t}^{8}} \right){{+}^{25}}{{C}_{5}}\left( {{t}^{10}} \right)+...............{{+}^{25}}{{C}_{25}}\left( {{t}^{50}} \right) \right\\} .
On multiplying we get, $^{25}{{C}_{5}}\left( {{t}^{50}} \right)$ .
In this case the coefficient of ${{t}^{50}}$ is $^{25}{{C}_{5}}$ ………………………………(10)
From equation (9) and equation (10), we have the coefficient of the term ${{t}^{50}}$ .
The coefficient of the term ${{t}^{50}}$ = $^{25}{{C}_{25}}{{+}^{25}}{{C}_{5}}=1{{+}^{25}}{{C}_{5}}$ .
Therefore, the coefficient of the term ${{t}^{50}}$ is $1{{+}^{25}}{{C}_{5}}$ .

So, the correct answer is “Option A”.

Note: In this question, one might think to simplify the expression ${{\left( 1+{{t}^{2}} \right)}^{25}}\left( 1+{{t}^{25}} \right)\left( 1+{{t}^{40}} \right)\left( 1+{{t}^{45}} \right)\left( 1+{{t}^{47}} \right)$ and then separate the term ${{t}^{50}}$ to obtain its coefficient. It is just too complex and next to impossible to multiply each term and simplify the expression with just pen and paper. So, we don’t have to approach this question by this method.