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JEE Mains Maths · Formula sheet

Binomial Theorem formulas

17 formulas and 17 common traps for JEE Mains Maths Binomial Theorem, grouped by subtopic.

Full notes with worked examples

The General Term

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Setting the power of x

General term

Tr+1=(nr) a n−r b rT_{r+1}=\binom nr\,a^{\,n-r}\,b^{\,r}

Simplify the bracket first

A common collapse

x+1x2/3−x1/3+1−x−1x−x=x1/3−x−1/2\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-\sqrt x}=x^{1/3}-x^{-1/2}

Equating coefficients from two expansions

Symmetry of coefficients

(nr)=(nn−r)\binom nr=\binom n{n-r}

Common traps

T with index r + 1

The term containing brb^r is the (r+1)(r+1)-th. 'The 7th term' means r=6r=6; using r=7r=7 shifts every answer.

Check the numerator's sign

x−1x−x\frac{x-1}{x-\sqrt x} collapses; x+1x−x\frac{x+1}{x-\sqrt x} does not. If the bracket will not simplify, re-read it before expanding a messy expression.

Signs from the second term

In (ax−1bx2)n\left(ax-\frac{1}{bx^2}\right)^{n} the term carries (−1)r(-1)^r. If the two coefficients have opposite signs, they cannot be equal for positive a,ba,b — check the parity of each rr.

Consecutive Coefficients and Special Terms

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The ratio of neighbouring coefficients

Neighbour ratio

(nr)(nr−1)=n−r+1r\frac{\binom nr}{\binom n{r-1}}=\frac{n-r+1}{r}

Middle, greatest and end terms

Start versus end

Tk (from start)Tk (from end)=(ab)n−2k+2\frac{T_k\ \text{(from start)}}{T_k\ \text{(from end)}}=\left(\frac ab\right)^{n-2k+2}

Common traps

Which r is which

Fix one labelling — say the middle coefficient is (nr)\binom nr — and write both ratios from it. Mixing rr for the first term in one ratio and for the middle in the other gives equations with no integer solution.

The exponent is n - 2k + 2

Between the kk-th term from the start and the kk-th from the end, the powers of aa differ by n−2k+2n-2k+2, not n−2kn-2k. Test with k=1k=1: the first and last terms differ by ana^n against bnb^n.

Products and Multinomial Expansions

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Coefficients in a product

Two factors

(1+x)p(1−x)q:[x]=p−q,[x2]=(p−q)2−(p+q)2(1+x)^p(1-x)^q:\quad[x]=p-q,\quad[x^2]=\tfrac{(p-q)^2-(p+q)}{2}

Three-term brackets

Multinomial term

(p+q+r)n:  n!a! b! c! pa qb rc,a+b+c=n(p+q+r)^n:\ \ \frac{n!}{a!\,b!\,c!}\,p^a\,q^b\,r^c,\quad a+b+c=n

Common traps

Include every combination

With a quadratic in front, the coefficient of xkx^k has three contributions, from x0x^0, x1x^1 and x2x^2. Dropping the last one is the usual slip.

List every triple

The two conditions usually allow several (a,b,c)(a,b,c). Write the free index's possible values in order before computing, so none is missed.

Rational and Integral Terms

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Counting and adding rational terms

Rational-term condition

(nr)pn−raqrb is rational exactly when a∣(n−r) and b∣r\binom nr p^{\frac{n-r}{a}}q^{\frac rb}\ \text{is rational exactly when}\ a\mid(n-r)\ \text{and}\ b\mid r

Integral parts through the conjugate

Conjugate pair

x=(a+b)n, y=(a−b)n:x+y∈Z,  0<y<1⇒[x]=x+y−1x=(a+\sqrt b)^n,\ y=(a-\sqrt b)^n:\quad x+y\in\mathbb Z,\ \ 0<y<1\Rightarrow[x]=x+y-1

Common traps

Both exponents must be whole

Checking only b∣rb\mid r is not enough: n−ra\frac{n-r}{a} must be an integer too. When aa does not divide nn, the valid rr are shifted, not multiples of lcm⁡(a,b)\operatorname{lcm}(a,b).

A negative conjugate

When a−ba-\sqrt b is negative, yy changes sign with nn. For odd nn, yy is negative and [x]=x+y[x]=x+y exactly; subtracting 1 by habit gives the wrong parity.

Coefficient Sums by Substitution and Differentiation

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Substituting x = 1, −1 and other values

Even and odd parts

∑r evenar=f(1)+f(−1)2,∑r oddar=f(1)−f(−1)2\sum_{r\ \text{even}}a_r=\frac{f(1)+f(-1)}{2},\qquad\sum_{r\ \text{odd}}a_r=\frac{f(1)-f(-1)}{2}

Weighted sums: r C(n, r) and r² C(n, r)

Weighted sums

∑rr(nr)=n 2n−1,∑rr2(nr)=n(n+1) 2n−2\sum_{r}r\binom nr=n\,2^{n-1},\qquad\sum_{r}r^2\binom nr=n(n+1)\,2^{n-2}

Common traps

Remove the terms outside the range

If the sum stops before the last odd coefficient (say at a37a_{37} out of a39a_{39}), compute the full odd sum and subtract the missing coefficient separately.

r² is not r · r in the identity

Applying r(nr)=n(n−1r−1)r\binom nr=n\binom{n-1}{r-1} twice needs the second rr rewritten as (r−1)+1(r-1)+1. Treating r2(nr)r^2\binom nr as n2(n−2r−2)n^2\binom{n-2}{r-2} is the standard wrong turn.

Sums with Fractions and Products of Coefficients

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Coefficients divided by r + 1

Absorption into n + 1

∑r=0n(nr)r+1=1n+1∑r=0n(n+1r+1)=2n+1−1n+1\sum_{r=0}^{n}\frac{\binom nr}{r+1}=\frac{1}{n+1}\sum_{r=0}^{n}\binom{n+1}{r+1}=\frac{2^{n+1}-1}{n+1}

Products of coefficients: Vandermonde

Vandermonde's identity

∑r(mr)(nk−r)=(m+nk)\sum_{r}\binom mr\binom n{k-r}=\binom{m+n}{k}

Common traps

Trim the ends you are not given

If the sum stops at (119)10\frac{\binom{11}{9}}{10}, the identity gives 112∑(12j)\frac{1}{12}\sum\binom{12}{j} over j=2,…,10j=2,\dots,10 only. Subtract the missing (120),(121),(1211),(1212)\binom{12}{0},\binom{12}{1},\binom{12}{11},\binom{12}{12} from 2122^{12}.

Match the indices before summing

∑(mr)(nr)\sum\binom mr\binom nr is not (m+nr)\binom{m+n}{r} — rr is the summation variable. Rewrite one factor so the lower indices add to a fixed number; that fixed number is the answer's lower index.

Sums of Expansions: Hockey Stick and Geometric Series

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The hockey-stick identity

Hockey stick

∑k=rn(kr)=(n+1r+1)\sum_{k=r}^{n}\binom kr=\binom{n+1}{r+1}

Summing a geometric series of expansions

Collapsing the sum

∑k=0nxk(1+x)n−k=(1+x)n+1−xn+1\sum_{k=0}^{n}x^k(1+x)^{n-k}=(1+x)^{n+1}-x^{n+1}

Common traps

Start the column at the right row

The identity needs the sum to start at (rr)\binom rr. If it starts at (mr)\binom mr with m>rm>r, subtract (mr+1)\binom m{r+1} — not (m−1r+1)\binom{m-1}{r+1} or (mr)\binom{m}{r}.

The top power goes up by one

Summing n+1n+1 terms produces (1+x)n+1(1+x)^{n+1}, not (1+x)n(1+x)^n. The coefficient of xrx^r is therefore (n+1r)\binom{n+1}{r}; using (nr)\binom nr is the usual error.

Remainders and Divisibility

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Remainders of large powers

Base near a multiple

(mq±1)n=m(… )+(±1)n(mq\pm1)^n=m(\dots)+(\pm1)^n

What an expression is divisible by

Removing the first two terms

(1+m)n−mn−1=m2[(n2)+(n3)m+… ](1+m)^n-mn-1=m^2\left[\binom n2+\binom n3m+\dots\right]

Common traps

Leftover factors

If the exponent is not a multiple of the cycle, a factor remains outside the bracket: 2100=2⋅8332^{100}=2\cdot8^{33}. Forgetting that factor gives remainder 1 instead of 2.

Pair the terms to match the signs

an−bn−cn+dna^n-b^n-c^n+d^n can be grouped as (an−cn)−(bn−dn)(a^n-c^n)-(b^n-d^n) or (an−bn)−(cn−dn)(a^n-b^n)-(c^n-d^n). Each grouping gives a different common factor; try both.

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