Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Binomial Theorem question

2022 · 25 Jun · Shift 1 · Q28
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Binomial Theorem
  5. /2022 · 25 Jun · Shift 1 · Q28

Binomial Theorem question

2022 · 25 Jun · Shift 1 · Q28

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
If 12 . 310+122 . 39+  .....  +  1210 . 3=K210 . 310{1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,.....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}2.3101​+22.391​+.....+210.31​=210.310K​, then the remainder when K is divided by 6 is :
  1. A
    1
  2. B
    2
  3. C
    3
  4. D
    5
View written solutionFree

Correct answer: D

  1. Write the sum in a simpler form

Given

S=12⋅310+122⋅39+⋯+1210⋅3S=\frac{1}{2\cdot 3^{10}}+\frac{1}{2^2\cdot 3^9}+\cdots+\frac{1}{2^{10}\cdot 3}S=2⋅3101​+22⋅391​+⋯+210⋅31​

The general term is

12r 311−r,r=1,2,…,10.\frac{1}{2^r\,3^{11-r}},\qquad r=1,2,\dots,10.2r311−r1​,r=1,2,…,10.

So

S=∑r=11012r311−r.S=\sum_{r=1}^{10}\frac{1}{2^r3^{11-r}}.S=r=1∑10​2r311−r1​.

We are told that

S=K210310.S=\frac{K}{2^{10}3^{10}}.S=210310K​.
  1. Convert each term to denominator 2103102^{10}3^{10}210310

For the rrr-th term,

12r311−r=210−r3r−1210310.\frac{1}{2^r3^{11-r}} =\frac{2^{10-r}3^{r-1}}{2^{10}3^{10}}.2r311−r1​=210310210−r3r−1​.

Hence

K=∑r=110210−r3r−1.K=\sum_{r=1}^{10}2^{10-r}3^{r-1}.K=r=1∑10​210−r3r−1.

Let j=r−1j=r-1j=r−1. Then j=0j=0j=0 to 999, so

K=∑j=0929−j3j.K=\sum_{j=0}^{9}2^{9-j}3^j.K=j=0∑9​29−j3j.

This is a geometric-type sum:

K=29+28⋅3+27⋅32+⋯+39.K=2^9+2^8\cdot 3+2^7\cdot 3^2+\cdots+3^9.K=29+28⋅3+27⋅32+⋯+39.
  1. Use the identity for mixed powers

Recall:

an+an−1b+⋯+bn=an+1−bn+1a−b.a^n+a^{n-1}b+\cdots+b^n=\frac{a^{n+1}-b^{n+1}}{a-b}.an+an−1b+⋯+bn=a−ban+1−bn+1​.

Here a=2a=2a=2, b=3b=3b=3, and n=9n=9n=9. Thus

K=210−3102−3=310−210.K=\frac{2^{10}-3^{10}}{2-3}=3^{10}-2^{10}.K=2−3210−310​=310−210.
  1. Find K mod 6K \bmod 6Kmod6

We need the remainder when KKK is divided by 666:

K=310−210.K=3^{10}-2^{10}.K=310−210.

Now,

  • 3103^{10}310 is divisible by 333, and for modulo 666, any power of 333 greater than 111 gives remainder 333: 310≡3(mod6).3^{10}\equiv 3 \pmod 6.310≡3(mod6).
  • For powers of 222, 21≡2,22≡4,23≡2,24≡4,…2^1\equiv 2,\quad 2^2\equiv 4,\quad 2^3\equiv 2,\quad 2^4\equiv 4,\dots21≡2,22≡4,23≡2,24≡4,… Since 101010 is even, 210≡4(mod6).2^{10}\equiv 4\pmod 6.210≡4(mod6).

Therefore,

K≡3−4≡−1≡5(mod6).K\equiv 3-4\equiv -1\equiv 5 \pmod 6.K≡3−4≡−1≡5(mod6).

So the remainder is

5.\boxed{5}.5​.
  1. Check with options

Option D\boxed{D}D​ is correct.

PreviousNext

More from Binomial Theorem

  • The coefficient of x101 in the expression (5+x)500+x(5+x)499+x2(5+x)498+.....+x500, x > 0, is2022 · MCQ
  • If the sum of the co-efficient of all the positive even powers of x in the binomial expansion of (2x3+x3​)10 is 510−β.39, then β is equal to ​.2022 · Numerical
  • The remainder when (2021)2023 is divided by 7 is :2022 · MCQ
  • The remainder when (2021)2022+(2022)2021 is divided by 7 is2022 · MCQ
  • Let for the 9th  term in the binomial expansion of (3+6x)n, in the increasing powers of 6x, to be the greatest for x=23​, the least value of n is n0​. If k is the…2022 · Numerical
  • If the coefficient of x10 in the binomial expansion of (541​x​​+x31​5​​)60 is 5k.l, where l, k ∈ N and l is co-prime to 5, then k is equal…2022 · Numerical
  • The remainder when 72022+32022 is divided by 5 is :2022 · MCQ
  • Let the coefficients of the middle terms in the expansion of (6​1​+βx)4,(1−3βx)2 and (1−2β​x)6,β>0, respectively form the first three terms of an A.P. If…2022 · Numerical