Question

#Maths Aryabhattas only: Find the remainder when
${(x + 1)}^{n}$
is divided by (x - 1)³ where x>3​

Answer 1

We are asked to find the remainder obtained on dividing $(x+1)^n$ by $(x-1)^3.$

We see that,

$\longrightarrow (x+1)^n=(x-1+2)^n$

Or,

$\longrightarrow (x+1)^n=((x-1)+2)^n$

On performing binomial expansion,

$\displaystyle\longrightarrow (x+1)^n=\sum_{r=0}^n^n\!C_r\,(x-1)^{n-r}\,2^r$

But we separately represent last few terms in the expansion as follows:

$\displaystyle\longrightarrow (x+1)^n=\sum_{r=0}^{n-3}^n\!C_r\,(x-1)^{n-r}\,2^r+\sum_{r=n-2}^{n}^n\!C_r\,(x-1)^{n-r}\,2^r$

On dividing each term by $(x-1)^3,$

$\displaystyle\longrightarrow\dfrac{(x+1)^n}{(x-1)^3}=\dfrac{\displaystyle\sum_{r=0}^{n-3}^n\!C_r\,(x-1)^{n-r}\,2^r+\sum_{r=n-2}^{n}^n\!C_r\,(x-1)^{n-r}\,2^r}{(x-1)^3}$

$\displaystyle\longrightarrow\dfrac{(x+1)^n}{(x-1)^3}=\sum_{r=0}^{n-3}^n\!C_r\,(x-1)^{n-r-3}\,2^r+\dfrac{\displaystyle\sum_{r=n-2}^{n}^n\!C_r\,(x-1)^{n-r}\,2^r}{(x-1)^3}$

From this, since $3\ \textgreater\ 2,\ 1,\ 0,$ we get the remainder,

$\displaystyle\longrightarrow r(x)=\sum_{r=n-2}^{n}^n\!C_r\,(x-1)^{n-r}\,2^r$

$\displaystyle\longrightarrow r(x)=\ ^n\!C_{n-2}\,(x-1)^2\,2^{n-2}+\ ^n\!C_{n-1}\,(x-1)\,2^{n-1}+\ ^n\!C_n\,2^{n}$

$\longrightarrow\underline{\underline{r(x)=\ ^n\!C_2\,(x-1)^2\,2^{n-2}+\ ^n\!C_1\,(x-1)\,2^{n-1}+2^{n}}}$

$\longrightarrow\underline{\underline{r(x)=n(n-1)\,(x-1)^2\,2^{n-3}+n(x-1)\,2^{n-1}+2^{n}}}$

Answer 2

please make me Brianlist and follow me