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NCERT Exemplar · Q8

Q.Find the sixth term of the expansion (y1/2+x1/3)n\left(y^{1/2} + x^{1/3}\right)^{n}, if the binomial coefficient of the third term from the end is 4545.

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The binomial coefficient of the third term from the end equals the coefficient of the third term from the beginning; setting (n2)=45\binom{n}{2} = 45 gives n=10n = 10. The sixth term is then (105)y5/2x5/3=252y5/2x5/3\binom{10}{5} y^{5/2} x^{5/3} = 252 y^{5/2} x^{5/3}.

Why this approach works

The Binomial Theorem tells us that (a+b)n=∑r=0n(nr)an−rbr(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r. Each term has a binomial coefficient (nr)\binom{n}{r} and powers that decrease in aa while increasing in bb.

A key symmetry property: the binomial coefficient of the kk-th term from the end equals the coefficient of the kk-th term from the beginning, because (nr)=(nn−r)\binom{n}{r} = \binom{n}{n-r}. This symmetry will let us find nn first, then compute the requested term.

Finding nn from the given condition

The expansion has n+1n+1 terms total (from r=0r=0 to r=nr=n).

  1. Identify the third term from the end.

    The last term corresponds to r=nr = n, the second-last to r=n−1r = n-1, so the third from the end corresponds to r=n−2r = n - 2.

    Its binomial coefficient is (nn−2)\binom{n}{n-2}.

  2. Use the symmetry property.

    We know (nn−2)=(n2)\binom{n}{n-2} = \binom{n}{2} by the symmetry of binomial coefficients.

    The problem states this coefficient equals 4545:

(n2)=45\binom{n}{2} = 45

  1. Solve for nn.

(n2)=n(n−1)2=45\binom{n}{2} = \frac{n(n-1)}{2} = 45

n(n−1)=90n(n-1) = 90

n2−n−90=0n^2 - n - 90 = 0

Factoring: (n−10)(n+9)=0(n-10)(n+9) = 0.

Since nn must be positive, n=10n = 10.

Tip

The symmetry (nr)=(nn−r)\binom{n}{r} = \binom{n}{n-r} means "third from the end" is the same as "third from the beginning" in terms of coefficients — a huge shortcut.

Finding the sixth term

Now we expand (y1/2+x1/3)10(y^{1/2} + x^{1/3})^{10} and find its sixth term.

  1. General term formula. …

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