Easy Practice — Set 1

Quick Recap — Binomial Theorem

• $(a+b)^n=\sum_{r=0}^{n}{}^nC_r\,a^{n-r}b^r$, with $n+1$ terms.
• General term: $T_{r+1}={}^nC_r\,a^{n-r}b^r$.
• Coefficients of $(1+x)^n$ are $^nC_0,{}^nC_1,\dots,{}^nC_n$; the row sum $=2^n$.
• First term $a^n$, last term $b^n$; coefficients are symmetric.

Beyond-NCERT JEE Formulae

Past the NCERT core (general term, middle term, the row sum $\sum {}^{n}C_{r}=2^{n}$, the square sum $\sum {}^{n}C_{r}^{2}={}^{2n}C_{n}$ and the greatest binomial coefficient are assumed known), these are the high-yield extras that decide JEE Main marks.

1. Numerically greatest term in $(1+x)^{n}$

Compare consecutive terms through their ratio: $$\dfrac{T_{r+1}}{T_{r}}=\dfrac{n-r+1}{r}\,|x|.$$ The terms increase while this ratio exceeds $1$ and decrease once it drops below $1$. Solving $\dfrac{n-r+1}{r}|x|\ge 1$ gives $r\le\dfrac{(n+1)|x|}{1+|x|}$, so the numerically greatest term is $T_{m+1}$ (the $r=m$ term) with $$m=\left\lfloor\dfrac{(n+1)|x|}{1+|x|}\right\rfloor.$$ [JEE Tip] If $\dfrac{(n+1)|x|}{1+|x|}$ is itself an integer, the ratio equals $1$ there, so $T_{m}$ and $T_{m+1}$ are equal and jointly greatest. For $(a+bx)^{n}$ first pull out $a^{n}$ to write $a^{n}\left(1+\dfrac{b}{a}x\right)^{n}$, then apply the rule with $\dfrac{b}{a}x$ in place of $x$.

2. Binomial series for any real index

For a real index $n$ (fractional or negative) and $|x|<1$, $$(1+x)^{n}=1+nx+\dfrac{n(n-1)}{2!}x^{2}+\dfrac{n(n-1)(n-2)}{3!}x^{3}+\cdots$$ This is an infinite series unless $n$ is a non-negative integer. For small $x$ the linear estimate $(1+x)^{n}\approx 1+nx$ is quick; keep the $x^{2}$ term for four-decimal accuracy. Useful special cases are $(1-x)^{-1}=1+x+x^{2}+\cdots$ and $(1+x)^{-2}=1-2x+3x^{2}-\cdots$.

3. Weighted coefficient sums

Differentiating or integrating $(1+x)^{n}$ and then setting $x=1$ produces $$\sum_{r=0}^{n} r\,{}^{n}C_{r}=n\,2^{n-1},\qquad \sum_{r=0}^{n} r^{2}\,{}^{n}C_{r}=n(n+1)\,2^{n-2},$$ $$\sum_{r=0}^{n}\dfrac{{}^{n}C_{r}}{r+1}=\dfrac{2^{n+1}-1}{n+1}.$$ The first two follow from $r\,{}^{n}C_{r}=n\,{}^{n-1}C_{r-1}$; the third from $\dfrac{{}^{n}C_{r}}{r+1}=\dfrac{1}{n+1}\,{}^{n+1}C_{r+1}$ (equivalently, integrate $(1+x)^{n}$ over $[0,1]$).

4. Number of rational (integral) terms

In $\left(a^{1/p}+b^{1/q}\right)^{n}$ the general term is $T_{r+1}={}^{n}C_{r}\,a^{(n-r)/p}\,b^{r/q}$. It is rational only when both exponents $\dfrac{n-r}{p}$ and $\dfrac{r}{q}$ are integers. [JEE Tip] The tighter condition $\dfrac{r}{q}\in\mathbb{Z}$ usually shortlists $r$ to $0,q,2q,\dots,n$; test each against $\dfrac{n-r}{p}\in\mathbb{Z}$ and count the survivors, which sit $\operatorname{lcm}(p,q)$ apart.

5. Multinomial expansion

For $k$ symbols, $$(x_{1}+x_{2}+\cdots+x_{k})^{n}=\sum \dfrac{n!}{n_{1}!\,n_{2}!\cdots n_{k}!}\,x_{1}^{n_{1}}x_{2}^{n_{2}}\cdots x_{k}^{n_{k}},$$ the sum running over all non-negative integers with $n_{1}+n_{2}+\cdots+n_{k}=n$. The number of distinct terms equals the number of such solutions, $$N={}^{\,n+k-1}C_{k-1}.$$ For $k=3$ this is ${}^{\,n+2}C_{2}=\dfrac{(n+1)(n+2)}{2}$, and the largest multinomial coefficient occurs when the $n_{i}$ are made as equal as possible.

Solved Examples — Beyond-NCERT Formulae

Each example names the formula it uses and ends with a quick check.

Example 1 — Numerically greatest term in $(1+x)^{n}$

Problem. Find the numerically greatest term in the expansion of $(1+x)^{10}$ when $x=\dfrac{2}{3}$.

Solution. With $n=10$ and $|x|=\dfrac{2}{3}$, $$\dfrac{(n+1)|x|}{1+|x|}=\dfrac{11\cdot\frac{2}{3}}{1+\frac{2}{3}}=\dfrac{22/3}{5/3}=\dfrac{22}{5}=4.4,$$ so $m=\lfloor 4.4\rfloor=4$ and the greatest term is $T_{5}$ (the $r=4$ term): $$T_{5}={}^{10}C_{4}\left(\dfrac{2}{3}\right)^{4}=210\cdot\dfrac{16}{81}=\dfrac{1120}{27}\approx 41.5.$$

Check. $T_{4}={}^{10}C_{3}\left(\frac{2}{3}\right)^{3}=\dfrac{960}{27}\approx 35.6$ and $T_{6}={}^{10}C_{5}\left(\frac{2}{3}\right)^{5}=\dfrac{8064}{243}\approx 33.2$, both smaller than $T_{5}$.

Answer. $T_{5}=\dfrac{1120}{27}$.

Example 2 — Greatest term after factoring, equal-terms case

Problem. Find the numerically greatest term in the expansion of $(2+3x)^{9}$ when $x=1$.

Solution. Pull out $2^{9}$: $(2+3x)^{9}=2^{9}\left(1+\dfrac{3}{2}x\right)^{9}$, so at $x=1$ the effective variable is $y=\dfrac{3}{2}$ with $n=9$. Then $$\dfrac{(n+1)|y|}{1+|y|}=\dfrac{10\cdot\frac{3}{2}}{1+\frac{3}{2}}=\dfrac{15}{5/2}=6,$$ an integer, so $T_{6}$ and $T_{7}$ are equal and jointly greatest. Using the original term $T_{r+1}={}^{9}C_{r}\,2^{9-r}\,3^{r}$, $$T_{6}={}^{9}C_{5}\,2^{4}\,3^{5}=126\cdot16\cdot243=489888,\qquad T_{7}={}^{9}C_{6}\,2^{3}\,3^{6}=84\cdot8\cdot729=489888.$$

Check. The two candidate terms coincide, exactly the integer-ratio tie predicted by the rule; every other term is strictly smaller.

Answer. The greatest term is $489888$, attained by both $T_{6}$ and $T_{7}$.

Example 3 — Approximation with an integer index

Problem. Estimate $(1.02)^{10}$ using the binomial series up to the $x^{3}$ term.

Solution. Write $1.02=1+0.02$ with $n=10$: $$(1+0.02)^{10}\approx 1+10(0.02)+\dfrac{10\cdot9}{2!}(0.02)^{2}+\dfrac{10\cdot9\cdot8}{3!}(0.02)^{3}.$$ Term by term this is $1+0.2+45(0.0004)+120(0.000008)=1+0.2+0.018+0.00096=1.21896$.

Check. The exact value is $(1.02)^{10}=1.218994\ldots$, so the four-term estimate is right to four decimals.

Answer. $(1.02)^{10}\approx 1.2190$.

Example 4 — Approximation with a fractional index

Problem. Estimate $\sqrt[3]{1.03}=(1.03)^{1/3}$ keeping terms up to $x^{2}$.

Solution. Here $n=\dfrac{1}{3}$ and $x=0.03$, so $$(1+x)^{1/3}\approx 1+\dfrac{1}{3}x+\dfrac{\frac{1}{3}\left(\frac{1}{3}-1\right)}{2!}x^{2}=1+\dfrac{1}{3}(0.03)+\dfrac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2}(0.03)^{2}.$$ This gives $1+0.01-\dfrac{1}{9}(0.0009)=1+0.01-0.0001=1.0099$.

Check. Cubing back, $(1.0099)^{3}=1.02999\ldots\approx 1.03$; the exact root is $1.009901\ldots$.

Answer. $\sqrt[3]{1.03}\approx 1.0099$.

Example 5 — Weighted sums $\sum r\,{}^{n}C_{r}$ and $\sum r^{2}\,{}^{n}C_{r}$

Problem. Evaluate $\displaystyle\sum_{r=0}^{10} r\,{}^{10}C_{r}$ and $\displaystyle\sum_{r=0}^{10} r^{2}\,{}^{10}C_{r}$.

Solution. Apply the closed forms with $n=10$: $$\sum_{r=0}^{10} r\,{}^{10}C_{r}=n\,2^{n-1}=10\cdot2^{9}=5120,$$ $$\sum_{r=0}^{10} r^{2}\,{}^{10}C_{r}=n(n+1)2^{n-2}=10\cdot11\cdot2^{8}=110\cdot256=28160.$$

Check. Summing the ten coefficients directly reproduces $5120$ and $28160$.

Answer. $5120$ and $28160$.

Example 6 — Sum weighted by $\dfrac{1}{r+1}$

Problem. Evaluate $\displaystyle\sum_{r=0}^{5}\dfrac{{}^{5}C_{r}}{r+1}$.

Solution. Use $\displaystyle\sum_{r=0}^{n}\dfrac{{}^{n}C_{r}}{r+1}=\dfrac{2^{n+1}-1}{n+1}$ with $n=5$: $$\sum_{r=0}^{5}\dfrac{{}^{5}C_{r}}{r+1}=\dfrac{2^{6}-1}{6}=\dfrac{63}{6}=\dfrac{21}{2}.$$

Check. Adding term by term, $1+\dfrac{5}{2}+\dfrac{10}{3}+\dfrac{10}{4}+\dfrac{5}{5}+\dfrac{1}{6}=\dfrac{21}{2}=10.5$.

Answer. $\dfrac{21}{2}$.

Example 7 — Number of rational terms

Problem. How many terms in the expansion of $\left(\sqrt{2}+\sqrt[3]{3}\right)^{12}$ are rational?

Solution. The general term is $$T_{r+1}={}^{12}C_{r}\,2^{(12-r)/2}\,3^{r/3}.$$ Rationality needs $\dfrac{12-r}{2}\in\mathbb{Z}$ (so $r$ is even) and $\dfrac{r}{3}\in\mathbb{Z}$ (so $3\mid r$). Both hold together exactly when $6\mid r$, giving $r=0,6,12$ — three values.

Check. $r=0:\ 2^{6}=64$; then $r=6:\ {}^{12}C_{6}\,2^{3}3^{2}=924\cdot72=66528$; then $r=12:\ 3^{4}=81$ — all rational, and no other $r$ qualifies.

Answer. $3$ rational terms.

Example 8 — Number of terms in a multinomial

Problem. How many distinct terms appear in the expansion of $(x+y+z+w)^{10}$ after like monomials are collected?

Solution. Here $k=4$ symbols and $n=10$, so the count is $$N={}^{\,n+k-1}C_{k-1}={}^{13}C_{3}=\dfrac{13\cdot12\cdot11}{6}=286.$$

Check. This is the number of non-negative integer solutions of $a+b+c+d=10$, which a stars-and-bars count confirms as $286$.

Answer. $286$ terms.