Question

The coefficient of $$x^{k}$$ in the expansion of
$$\mathrm{E} =1+\left( 1+x\right) +\left( 1+x\right)^{2} +\ldots +\left( 1+x\right)^{n} $$ is:(n>k)
(A). $${}^{n}C_{k}$$
(B). $${}^{n+1}C_{k}$$
(C). $${}^{n+1}C_{k+1}$$
(D). None of these

Answer 1

Hint: In this question it is given that we have to find the coefficient of $$x^{k}$$ from the expansion $$\mathrm{E} =1+\left( 1+x\right) +\left( 1+x\right)^{2} +\ldots +\left( 1+x\right)^{n} $$. So to find the solution we need to know the binomial expansion of $$\left(1+b\right)^{n} $$, which is,
$$\left(1+b\right)^{n} =\ ^{n} C_{0}+\ ^{n} C_{1}b+\ ^{n} C_{2}b^{2}+\ldots +\ ^{n} C_{n}b^{n}$$

Complete step-by-step solution:
Given expansion,
$$\mathrm{E} =1+\left( 1+x\right) +\left( 1+x\right)^{2} +\ldots +\left( 1+x\right)^{n} $$
The above equation is in the form of G.P(Geometric Progression) where first term is a=1 and the common ratio is r=(1+x) and total number of terms are (n+1),
And as we know that the sum of first (n+1) terms in a G.P ,
$$\mathrm{E} =a\cdot \dfrac{r^{n+1}-1}{r-1}$$
$$=1\cdot \dfrac{\left( 1+x\right)^{n+1} -1}{\left( 1+x\right) -1}$$
$$\mathrm{E} =\dfrac{\left( 1+x\right)^{n+1} -1}{x}$$
$$=\dfrac{\left( 1+x\right)^{n+1} }{x} -\dfrac{1}{x}$$ .......(1)
Now since the binomial term $$\left( 1+x\right)^{n+1}$$ is divided by x then we have to find its $$(k+1)^{th}$$ term,
$$\therefore \left( 1+x\right)^{n+1} =\ ^{n+1} C_{0}+\ ^{n+1} C_{1}\ x+\ldots +\ ^{n+1} C_{k+1}\ x^{k+1}+\ldots +\ ^{n+1} C_{n+1}\ x^{n+1}$$.........(2)
Putting the value of equation (2) in equation (1) , we get,
$$\therefore \mathrm{E} =\ ^{n+1} C_{0}\dfrac{1}{x} +\ ^{n+1} C_{1}\ \dfrac{x}{x} +\ldots +\ ^{n+1} C_{k+1}\ \dfrac{x^{k+1}}{x} +\ldots +\ ^{n+1} C_{n+1}\ \dfrac{x^{n+1}}{x} -\dfrac{1}{x}$$
$$=\ ^{n+1} C_{0}\dfrac{1}{x} +\ ^{n+1} C_{1}+\ldots +\ ^{n+1} C_{k+1}\ x^{k}+\ldots +\ ^{n+1} C_{n+1}\ x^{n}-\dfrac{1}{x}$$
Therefore, from the above we can say that the coefficient of $$x^{k}$$ is $${}^{n+1}C_{k+1}$$
Hence the correct option is option C.

Note: While solving we have directly used the formula of G.P, so to identify any series are in the form of G.P or not, then you have to observe that if any series is in the form of $$a+ar+ar^{2}+\ldots +ar^{n-1}$$ where n number of terms are there, then this series is in G.P where first term is a and the common ratio is r, i.e the ratio of the two consecutive terms is equal to r. Then the sum of the n term is $$S_{n}=a\dfrac{r^{n}-1}{r-1}$$,
But in the series the total number of terms was (n+1) so that's why we used (n+1) instead of n.