The coefficient of $x^{101}$ in the expansion of the expression $(5+x)^{500}+x(5+x)^{499}....+x^{500}$ is

Question

The coefficient of $x^{101}$ in the expansion of the expression $$(5+x)^{500}+ x(5+x)^{499}+x^2(5+x)^{498}+\cdots+x^{500}$$ is?

My attempt:- The coefficient of $x^{101}$ in the first term is $C(500,102)$,the second term is $C(499,101)$ and so on

so the general coefficient of the term with $x^{101}$ is $C(500-r,102-r)$ .

My question is how do I add these coefficients up?

source :- Jee mains 2022

Answer 1

Note that the given expression is $$\frac{(5+x)^{501}-x^{501}}{5}$$

So we can calculate the coefficient of $x^{101}$ of this expression which is $$5^{399} \cdot \binom{501}{101}$$ by the Binomial formula.

Note: To see how the given expression is simplified, you can notice that it is the same thing as $$\sum_{r=0}^{500} x^{500-r}(5+x)^{r} =x^{500}\cdot \sum_{r=0}^{500} \left(\frac{5+x}{x}\right)^r$$ which is easily summed as a geometric sum. More generally, $$\sum_{r=0}^n a^{n-r}b^r = \frac{a^{n+1}-b^{n+1}}{a-b}$$

Answer 2

Solution : Observe that by Properties of Geometric Progression (G.P.) $$(5+x)^{500}+x.(5+x)^{499}+ x^2.(5+x)^{498} +\cdots + x^{500}$$ $$=\frac{(5+x)^{501}-x^{501}}{(5+x)-x}$$ $$=\frac{(5+x)^{501}-x^{501}}{5}$$ Therefore by Bionomial Theorem the coefficient of $x^{101}$ $$=\frac{\binom{501}{101}.5^{400}}{5}=\boxed{\binom{501}{101}.5^{399}}$$

Answer 3

COMMENT.-According to development of $\dfrac{a^n-b^n}{a-b}$one has $$(5+x)^{500}+ x(5+x)^{499}+x^2(5+x)^{498}+\cdots+x^{500}=\dfrac{(5+x)^{501}-x^{501}}{5}$$ the numerator is equal to $$5^{501}+\binom{501}{1}5^{500}x+\cdots+\binom{501}{1}5x^{500}$$ and the required coefficient is $$\dfrac{\binom{501}{101}5^{400}}{5}=\frac{5^{399}(501!)}{(101!)(400!)}$$ It is a big number. For sake of fun, remember that if $p$ is a prime least to $n$ then the exponent of $p$ in $n!$ is equal to $$\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^2}\right\rfloor+\left\lfloor\frac{n}{p^3}\right\rfloor+\cdots$$ With this we can get rightfordwardly $$501!=2^{494}\cdot3^{248}\cdot5^{124}\cdot7^{82}\cdot11^{49}\cdot13^{40}\cdot17^{30}\cdot19^{27}\cdot23\cdot29\cdot31\cdots499$$ (all the successive primes from $23$ till $499$ with exponent $1$ THIS IS NOT TRUE).

CORRECTION.- The value of $501!$ It's much bigger than what I wrote down. I had a lapse in calculating the exponents for primes between $23$ and $499$ because each of these primes has a square greater than $501$. In truth, the calculated value should have been: $$501!=2^{494}\cdot3^{248}\cdot5^{124}\cdot7^{82}\cdot11^{49}\cdot13^{40}\cdot17^{30}\cdot19^{27}\cdot23^{21}\cdot29^{17}\cdot31^{16}\cdot37^{13}\cdot41^{12}\cdot43^{11}\cdot47^{10}\cdot53^9\cdot(59\cdot61)^8(67\cdot71)^7(73\cdot79\cdot83)^6(89\cdot97)^5(101\cdot103\cdot107\cdot109\cdot113)^4(127\cdots167)^3(173\cdots241)^2(251\cdots499)$$ In similar way, the calculation of $101!$ and $400!$ allows us to write the required coefficient in its decomposition in prime factors. The result would be an evident very big number.

Answer 4

The coefficient will be $\displaystyle 5^{399} \left[\binom{500}{399}+\binom{499}{399}+\ldots+\binom{399}{399} \right] = 5^{399} \binom{501}{400} =5^{399} \binom{501}{101} $ by Hockey Stick Identity