Let the smallest value of k N, for which the coefficient of x^3 in (1+x)^3+(1+x)^4+(1+x)^5…

Question

Let the smallest value of $k\in N$, for which the coefficient of ${x}^{3}$ in $(1+x{)}^{3}+(1+x{)}^{4}+(1+x{)}^{5}+\ldots +(1+x{)}^{99}+(1+kx{)}^{100},x
e 0$, is $(43n+\frac{101}{4})({}^{100}{C}_{3})$ for some $n\in N$, be $p$. Then the value of $p+n$ is :

Accepted Answer

Coefficient of ${x}^{3}$ in the given expression

$(1+x{)}^{3}+(1+x{)}^{4}+\cdots +(1+x{)}^{99}+(1+kx{)}^{100}$

is obtained by adding the coefficients of ${x}^{3}$ from each term.

For $(1+x{)}^{r}$, coefficient of ${x}^{3}$ is

${}^{r}{C}_{3}$

So, from

$(1+x{)}^{3}+(1+x{)}^{4}+\cdots +(1+x{)}^{99}$

the total coefficient of ${x}^{3}$ is

$\sum _{99}^{r=3}{}^{r}{C}_{3}$

Also, in $(1+kx{)}^{100}$, coefficient of ${x}^{3}$ is

${}^{100}{C}_{3}\, {k}^{3}$

Hence total coefficient of ${x}^{3}$ is

$\sum _{99}^{r=3}{}^{r}{C}_{3}+{}^{100}{C}_{3}{k}^{3}$

Now use the standard NCERT identity:

$\sum _{99}^{r=3}{}^{r}{C}_{3}={}^{100}{C}_{4}$

So coefficient becomes

${}^{100}{C}_{4}+{}^{100}{C}_{3}{k}^{3}$

Given that this equals

$(43n+\frac{101}{4}){}^{100}{C}_{3}$

Therefore,

${}^{100}{C}_{4}+{}^{100}{C}_{3}{k}^{3}=(43n+\frac{101}{4}){}^{100}{C}_{3}$

Divide by ${}^{100}{C}_{3}$:

$\frac{{}^{100}{C}_{4}}{{}^{100}{C}_{3}}+{k}^{3}=43n+\frac{101}{4}$

Now,

$\frac{{}^{100}{C}_{4}}{{}^{100}{C}_{3}}=\frac{100-3}{4}=\frac{97}{4}$

So,

${k}^{3}+\frac{97}{4}=43n+\frac{101}{4}$

Thus,

${k}^{3}=43n+1$

We need the smallest $k\in N$ such that

${k}^{3}-1$

is divisible by $43$.

That is,

${k}^{3}\equiv 1\, (\bmod \, 43)$

Now check smallest natural numbers:

${1}^{3}=1\equiv 1\, (\bmod \, 43)$

So the smallest value is

$k=1$

Hence,

$1=43n+1$

which gives

$n=0$

But in the question, $n\in N$. In school-level convention, if $N={0,1,2,\ldots }$, then this is allowed.

Thus,

$p=1,\, n=0$

and

$p+n=1$

But this is not among the options, so clearly the intended convention is $n\ge 1$.

Now take the smallest $n\in N$ with $n\ge 1$ such that

${k}^{3}=43n+1$

Try $n=1$:

${k}^{3}=44\,$

$n=2$:

${k}^{3}=87\,$

$n=3$:

${k}^{3}=130\,$

$n=4$:

${k}^{3}=173$

$n=5$:

${k}^{3}=216={6}^{3}$

So smallest such $k$ is

$p=6,\, n=5$

Therefore,

$p+n=6+5=11$

Hence the correct answer is:

$11$

So, Option B is correct.

Solutions · 1