PYQ Vault

JEE Mains Maths · Binomial Theorem

Sums of Expansions: Hockey Stick and Geometric Series

Finding a coefficient in a sum of many expansions — (1 + x)^3 + (1 + x)^4 + … or (1 + x)^n + x(1 + x)^(n − 1) + … — either by the hockey-stick identity or by summing the geometric series first.

All Binomial Theorem notesJEE Mains Maths strategy

Why this matters

Eleven PYQs, three of them numerical answer. A sum of consecutive binomial coefficients down one column collapses to a single coefficient, and a sum of expansions whose ratio is fixed is a geometric series with a two-term closed form. Two ideas cover the page.

Concept 1 of 2: The hockey-stick identity

Adding one column of Pascal's triangle gives the entry one row down and one place right: (rr)+(r+1r)+⋯+(nr)=(n+1r+1)\binom rr+\binom{r+1}r+\dots+\binom nr=\binom{n+1}{r+1}. So the coefficient of xrx^r in (1+x)r+(1+x)r+1+⋯+(1+x)n(1+x)^r+(1+x)^{r+1}+\dots+(1+x)^n is (n+1r+1)\binom{n+1}{r+1}. A sum that starts later is a difference of two such totals.

Definition

  • ∑k=rn(kr)=(n+1r+1)\sum_{k=r}^{n}\binom kr=\binom{n+1}{r+1}.
  • ∑k=mn(kr)=(n+1r+1)−(mr+1)\sum_{k=m}^{n}\binom kr=\binom{n+1}{r+1}-\binom{m}{r+1}.
  • (m+kk)=(m+km)\binom{m+k}{k}=\binom{m+k}{m}: a diagonal sum becomes a column sum.

Hockey stick

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

Worked example

Find (102)+(112)+⋯+(152)\binom{10}{2}+\binom{11}{2}+\dots+\binom{15}{2}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 25 Jan 2023 · Q156Moderate

Example 1 · Binomial Theorem · Sums of Expansions: Hockey Stick and Geometric Series

∑k=0645+kC3\sum_{k=0}^{6} {}^{45+k}C_{3} is equal to

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}.
Drill 4 more on the hockey-stick identity

Concept 2 of 2: Summing a geometric series of expansions

In (1+x)n+x(1+x)n−1+⋯+xn(1+x)^n+x(1+x)^{n-1}+\dots+x^n, each term is the previous one times x1+x\frac{x}{1+x}. The geometric-series formula collapses it to (1+x)n+1−xn+1(1+x)^{n+1}-x^{n+1}, and any coefficient is then one binomial coefficient. The same idea with weights, ∑ktk\sum kt^k, uses the arithmetico-geometric sum.

Definition

  • ∑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}.
  • ∑k=0nxk(a+x)n−k=(a+x)n+1−xn+1a\sum_{k=0}^{n}x^k(a+x)^{n-k}=\frac{(a+x)^{n+1}-x^{n+1}}{a}.
  • ∑k=0n−1(x+3)n−1−k(x+2)k=(x+3)n−(x+2)n\sum_{k=0}^{n-1}(x+3)^{n-1-k}(x+2)^k=(x+3)^n-(x+2)^n.
  • ∑k=1nktk=t(1−(n+1)tn+ntn+1)(1−t)2\sum_{k=1}^{n}kt^k=\frac{t\left(1-(n+1)t^n+nt^{n+1}\right)}{(1-t)^2}.

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}

Worked example

Find the coefficient of x3x^{3} in (1+x)6+x(1+x)5+⋯+x6(1+x)^{6}+x(1+x)^{5}+\dots+x^{6}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 30 January 2023 · Q77Moderate

Example 2 · Binomial Theorem · Sums of Expansions: Hockey Stick and Geometric Series

The coefficient of x301x^{301} in (1+x)500+x(1+x)499+x2(1+x)498+……..x500(1 + x)^{500}+ x(1 + x)^{499}+x^{2}(1 + x)^{498}+ \ldots\ldots..x^{500} is :

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.
Drill 5 more on summing a geometric series of expansions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Watch out for (2)

Test yourself on Binomial Theorem

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Binomial Theorem chapter testDrill all 11 questions