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Calculation of remainder when $(x+1)^n$ is divided by $(x-1)^3$, where $n\in \mathbb{N}$

$\bf{My\; Try}::$ Using Division Algorithm:: $p(x) = q(x)\cdot g(x)+r(x)$

Now Let $r(x) = ax^2+bx+c$

So $(x+1)^n=q(x)\cdot (x-1)^3+ax^2+bx+c........................(1)$

Now put $x=1$, we get $\boxed{a+b+c=2^n}$

Now differentitae $(1)$, we get $n(x+1)^{n-1} = q(x)\cdot 3(x-1)^2+(x-1)^3\cdot q^{'}(x)+2ax+b$

again put $x=1$, we get $\boxed{2a+b=n(2)^{n-1}}$

Now again differentitae $(1)$ and then put $x=1$, we get

$\displaystyle \boxed{2a=n(n-1)2^{n-2}\Rightarrow \displaystyle a=\frac{n(n-1)}{2}\cdot 2^{n-2}}$

Similarly we get $\displaystyle b = n(2)^{n-1}-n(n-1)\cdot 2^{n-2}$

Similarly we get $\displaystyle c= 2^{n}+n(n-1)\cdot 2^{n-2}-\frac{n(n-1)}{2}\cdot 2^{n-2}$

So Remainder

$\displaystyle r(x) = \frac{n(n-1)}{2}2^{n-2}x^2+\left\{n(2)^{n-1}-n(n-1) 2^{n-2}\right\}x+2^{n}+n(n-1)\cdot 2^{n-2}-\frac{n(n-1)}{2}2^{n-2}$

is my solution is right , if not then how can i calculate it

or If there is any better method , then plz explain here

Thanks

juantheron
  • 53,015

3 Answers3

6

Another way : for integer $n\ge0$

$$(x+1)^n=(x-1+2)^n$$ $$\equiv2^n+\binom n12^{n-1}(x-1)+\binom n22^{n-2}(x-1)^2\left(\text{mod}{(x-1)^3}\right)$$

Clearly, the remainder reduces to $(x+1)^n$ for $0\le n\le2$

lab bhattacharjee
  • 274,582
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Your method is correct. You are essentially computing the first few terms of the Taylor power series expansion of $\,(x+1)^n$ at $\,x=1$ (or, equivalenly, of $\,(z+2)^n$ at $\,z =0,\,$ where $\,z = x-1).\,$ Alternatively, one may compute the expansion by using the Binomial Theorem.

Bill Dubuque
  • 272,048
2

Your method is correct. I have not checked your calculations but you can try some small values of $n$. At minimum check $n=0,1,2$ where you know that $q(x)=0$.

Is there a simpler method? I prefer to write

$$ (x+1)^n = q(x) (x-1)^3 + a \, (x-1)^2 + b (x-1) + c $$ and the proceed as you did. You can always expand out the final answer if you so choose. Try it and you will be amazed at how simple the calculations are.

Once you have your $a$, $b$ and $c$ you can always write $$ r=a x^2 + (b-2a) x + a - b + c$$ for the remainder $r$.

Try it, it is like magic!

user44197
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