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Mathematical Induction and Binomial Theorem question

2018 · Shift 2 · Q32
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  5. /2018 · Shift 2 · Q32

Mathematical Induction and Binomial Theorem question

2018 · Shift 2 · Q32

JEE AdvancedMathematicsMathematical Induction and Binomial TheoremNumerical+3 / −1
Let X=(10C1)2+2(10C2)2+3(10C3)2+...+10(10C10)2X = {({}^{10}{C_1})^2} + 2{({}^{10}{C_2})^2} + 3{({}^{10}{C_3})^2} + ... + 10{({}^{10}{C_{10}})^2}X=(10C1​)2+2(10C2​)2+3(10C3​)2+...+10(10C10​)2, where 10Cr{}^{10}{C_r}10Cr​, r ∈\in∈{1, 2, ..., 10} denote binomial coefficients. Then, the value of 11430X{1 \over {1430}}X14301​X is ..........
Numerical answer
View written solutionFree

Correct answer: 646

  1. We need to compute
X=∑r=110r(10r)2.X=\sum_{r=1}^{10} r\binom{10}{r}^2.X=r=1∑10​r(r10​)2.

Then find

X1430.\frac{X}{1430}.1430X​.
  1. Use the standard identity
r(nr)=n(n−1r−1).r\binom{n}{r}=n\binom{n-1}{r-1}.r(rn​)=n(r−1n−1​).

For n=10n=10n=10, this gives

r(10r)=10(9r−1).r\binom{10}{r}=10\binom{9}{r-1}.r(r10​)=10(r−19​).

Hence

r(10r)2=(10r)(r(10r))=10(10r)(9r−1).r\binom{10}{r}^2=\binom{10}{r}\left(r\binom{10}{r}\right) =10\binom{10}{r}\binom{9}{r-1}.r(r10​)2=(r10​)(r(r10​))=10(r10​)(r−19​).

So

X=10∑r=110(10r)(9r−1).X=10\sum_{r=1}^{10}\binom{10}{r}\binom{9}{r-1}.X=10r=1∑10​(r10​)(r−19​).
  1. Rewrite the index by putting k=r−1k=r-1k=r−1. Then k=0k=0k=0 to 999:
X=10∑k=09(10k+1)(9k).X=10\sum_{k=0}^{9}\binom{10}{k+1}\binom{9}{k}.X=10k=0∑9​(k+110​)(k9​).

Now use symmetry:

(9k)=(99−k).\binom{9}{k}=\binom{9}{9-k}.(k9​)=(9−k9​).

Thus

X=10∑k=09(10k+1)(99−k).X=10\sum_{k=0}^{9}\binom{10}{k+1}\binom{9}{9-k}.X=10k=0∑9​(k+110​)(9−k9​).

Let j=k+1j=k+1j=k+1. Then j=1j=1j=1 to 101010, so

X=10∑j=110(10j)(910−j).X=10\sum_{j=1}^{10}\binom{10}{j}\binom{9}{10-j}.X=10j=1∑10​(j10​)(10−j9​).
  1. Apply Vandermonde's identity:
∑j(mj)(nr−j)=(m+nr).\sum_{j} \binom{m}{j}\binom{n}{r-j}=\binom{m+n}{r}.j∑​(jm​)(r−jn​)=(rm+n​).

Here m=10m=10m=10, n=9n=9n=9, r=10r=10r=10. Therefore

∑j=110(10j)(910−j)=(1910).\sum_{j=1}^{10}\binom{10}{j}\binom{9}{10-j}=\binom{19}{10}.j=1∑10​(j10​)(10−j9​)=(1019​).

(There is no issue with limits because terms outside valid ranges are zero.)

So,

X=10(1910).X=10\binom{19}{10}.X=10(1019​).
  1. Compute (1910)\binom{19}{10}(1019​):
(1910)=(199)=92378.\binom{19}{10}=\binom{19}{9}=92378.(1019​)=(919​)=92378.

Thus

X=10×92378=923780.X=10\times 92378=923780.X=10×92378=923780.
  1. Now divide by 143014301430:
X1430=9237801430=646.\frac{X}{1430}=\frac{923780}{1430}=646.1430X​=1430923780​=646.
  1. Final answer:
646\boxed{646}646​
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