Binomial Theorem: questions with answers

Every surviving coefficient doubles, whichever term it belongs to — so the whole difference is 2 times the odd part of the expansion. Here that turns 56 into 112.

The minus is a property of the second QUANTITY, which is -3, and it is raised along with everything else. That is what makes the signs alternate all the way along instead of appearing once. If the minus really did belong to the bracket, (100 - 3)^6 would come out smaller than it does at every even index.

Each term carries a DIFFERENT power of B, and a power of a two-term bracket is itself a binomial expansion. Expanding B once and reusing it treats B² as though it were B, which it is not — and the higher the outer index, the more of the answer that mistake loses.

The binomial coefficients peak in the middle, but they are only half the story here: the other factor is a power of 5, and it climbs all the way to the last term. With 5 this large the two effects do not balance — the row's biggest entry, 6, sits on a term worth only 150, while the last term is 625. In (1 + x)⁴ the answer really would be 6, which is why the habit survives.

The constant sum is doing the work here rather than being recited. A learner who thinks the sum varies has nothing to subtract from and reaches for the index itself; one who thinks both indices move together copies 1.

The triangle grows downwards: row 9 is built from row 8, never the other way about. Reading the rule upwards turns the addition into a subtraction and gives 20, a number that is smaller than the entry above it -- impossible in the interior of the triangle, where every entry exceeds both of its parents.

Position and lower index differ by exactly one, always in the same direction. Answering 4 reads the lower index as though it were the count, and would place 6C0 at position 0 -- a position no row has.

The second term is NEGATIVE, because the second quantity is. Adding the three sizes as though they all pushed the same way gives a number above 1, and 0.95 raised to any power has to be below 1 — the answer refutes itself before any arithmetic is checked. Dropping the third term instead gives 0.7000, which is right as far as it goes but is not what was asked for.

Rewriting the base is not free: squaring the binomial is what the three-term form was hiding, so the index DOUBLES. Keeping the printed index carries the trinomial's own exponent onto a binomial that is smaller than it, and every later count is then wrong by a factor of two.

The coefficient is written next to a, which is what invites the pairing to be read left to right. But the lower index belongs to b: nC3 goes with b^3. The check is the index sum, 5 + 3 = 8; the swapped term happens to pass that check, which is why the pairing has to be known rather than reconstructed.

This is the degenerate case that shows why the chapter restricts itself. The expansion of $(1+x)^n$ owes its $n+1$ terms to the row of Pascal's triangle for index $n$, and there is no row for a negative index. An expansion for a negative index does exist in later work, but it is an unending series, which is a different object from the finite sum proved here.

Two steps, and the first is the one that gets skipped: 11 entries means index 10, and it is the INDEX that goes into the power of two. Using the count itself gives 2048, twice too much, and squaring the index gives 100.

The minus belongs to the second QUANTITY, not to the expansion, and squaring it cancels it. Distributing the bracket's minus over all 6 terms — the way one removes a bracket in front of a sum — would make this term -10, and would leave the expansion with no alternation at all.

The x in the denominator carries a NEGATIVE power, so the two exponents combine by subtraction even though the quantities are multiplied. Treating the denominator's power as positive gives 10, which is 4 too high, and the whole expansion then shows the wrong powers.

There are 200001 terms in that expansion and not one of them has to be written down. A comparison is a question about SIZE, and two terms already carry 2001; the index being enormous is what makes the second term big, not what makes the question unanswerable.