Concepts: general term, middle term and applications

Binomial Theorem24 min readLesson 3 of 5

3.1 The general term does the work Writing the general term of (a plus b) to the n as nCr times a to the (n minus r) times b to the r turns a term question into an equation. Simplify the power of x in that term, set it equal to the power the question asks for, and solve for r. If the solution is a whole number between 0 and n, the required term exists. This single method answers the standard forms: the coefficient of a particular power, the term independent of x, and the coefficient of the highest power. It also answers the reverse question, where a coefficient is given and the power or the constant is unknown. In every case, write the general term and simplify before computing any binomial coefficient. For the expansion of (2x squared plus 1 by x) to the tenth, the general term has x to the power 20 minus 3r. Setting that equal to 2 gives r equal to 6, and the coefficient is 10C6 times 2 to the fourth, which is 210 times 16 equal to 3360. For a term independent of x, set the power to 0 instead.

T_{r + 1} =^{n} C_{r} (2 x^{2})^{10 - r} (\frac{1}{x})^{r}, x^{20 - 3 r} = x^{2} \Rightarrow r = 6

Using the general term to find a required power

3.2 Middle terms and symmetric coefficients When n is even, the middle term is at position n by 2 plus 1, where the powers of the two quantities are equal. When n is odd, two terms share the middle positions, at (n plus 1) by 2 and (n plus 3) by 2, and their coefficients are equal because of the symmetry nCr equal to nC(n minus r). Symmetry answers many questions without computing. The coefficient of the term with b to the power r equals the coefficient of the term with b to the power n minus r, so a question that gives two equal coefficients can be turned into an equation for the unknown inside the bracket. This is the standard method when the middle-term coefficients of two different expansions are declared equal. The middle term also carries the largest coefficient in a binomial expansion, which is why questions ask where the greatest term occurs. In (1 plus x) to the n with positive x, the terms grow until the middle and then shrink, so the largest coefficient sits in the middle.

3.3 Sums of coefficients Substituting x equal to 1 in an expansion of the form (1 plus x) to the n gives the sum of all coefficients. Substituting x equal to minus 1 gives the alternating sum, which is 0, so the even-position coefficients and the odd-position coefficients have equal sums and each equals half of 2 to the n. For expansions with a constant c inside, the substitution changes. The coefficients of (1 plus cx) to the n add to (1 plus c) to the n. The coefficients of a difference add to the same value with c replaced by minus c. When a question multiplies two expansions or adds several expansions, differentiate or combine the sums rather than expanding term by term. A common question asks for the sum of coefficients of terms with only even powers. Add the expansions at x equal to 1 and x equal to minus 1, then divide by 2. This trick also isolates a single parity in a product of two binomials, which appears in the exercise set.

(1 + 1)^{n} = 2^{n} = r = 0 \sum n^{n} C_{r}, (1 - 1)^{n} = 0 = r = 0 \sum n (- 1)^{r}^{n} C_{r}

The two coefficient sums used most often

3.4 Approximations and divisibility The first few terms of an expansion give a usable approximation when the second quantity is small. For (0.99) to the fifth, write it as (1 minus 0.01) to the fifth and keep three terms: 1 minus 0.05 plus 0.001, which is 0.951, and the remaining terms are tiny. The sign pattern matters, because dropping a negative term makes the approximation larger than the exact value. Divisibility uses the same expansion with a base of the form 1 plus a. To show that 9 to the (n plus 1) minus 8n minus 9 is divisible by 64, expand (1 plus 8) to the (n plus 1) and collect the terms; after cancelling the constant and the linear term, every remaining term contains a factor of 64. The same pattern proves that 6 to the n minus 5n leaves remainder 1 on division by 25. These questions reward writing the base as 1 plus a first. Choosing the right a makes the modulus appear in the expansion; choosing the wrong one leaves a mess.

3.5 Exercise 7.1 and the miscellaneous exercise Exercise 7.1 asks for five expansions, including (1 minus 2x) to the fifth and (2x minus 3) to the sixth. It then asks for numerical values such as 96 cubed and 102 to the fifth, which are done by writing the number as 100 plus or minus a small number. The later questions ask for (a plus b) to the fourth minus (a minus b) to the fourth and for the sum (x plus 1) to the sixth plus (x minus 1) to the sixth, both of which are easiest by adding the two expansions and cancelling the odd terms. The divisibility questions ask for proofs that certain expressions are multiples of 64 and for the sum identity involving 3 to the r times nCr. The miscellaneous exercise adds the factorisation of a to the n minus b to the n, an approximation of (0.99) to the fifth using three terms, and the expansion of (3x squared minus 2ax plus 3a squared) cubed, which is first treated as a binomial in two terms.

Tip

For a numerical power such as 96 cubed, choose the split that gives a small second term: 96 as 100 minus 4, not 90 plus 6. The smaller the second term, the fewer terms you need.

in hinglish

Is page ka main tool general term hai. T_{r+1} = nCr a^{n−r}b^r likho, x ki power simplify karo, aur jo power chahiye uske barabar rakh kar r nikalo. Agar r whole number 0 se n ke beech aata hai to term exist karta hai; warna coefficient zero hai. Term independent of x ke liye power ko 0 karo. Middle term even n me n/2 + 1 par hota hai, odd n me do middle terms hote hain. Coefficient sums ke liye x = 1 rakho to 2^n milta hai, aur x = −1 se alternating sum 0; even powers alag karne ke liye dono ko add karke 2 se divide karo. Approximation me chhota second term chuno, jaise 96 = 100 − 4. Divisibility me base ko 1 + a likho. Har question me pehle general term, phir arithmetic. Final line me coefficient ka value aur uski condition dono likho, kyunki method marks ke saath saath answer ka format bhi check hota hai. Inhe revise karte waqt ek example khud solve karo, phir usi method se exercise ke sawal karo, kyunki general term aur middle term wale questions me sirf formula yaad rakhna kaafi nahi hota.