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Question: Match the Column: | | List I ...

Match the Column:

List IList II
(P)nC0+nC1−nC2−nC3+nC4+nC5−.......^nC_0 + ^nC_1 - ^nC_2 - ^nC_3 + ^nC_4 + ^nC_5 - .......(1)2n/2(2n/2−2+12cosnπ4)2^{n/2} \left( 2^{n/2-2} + \frac{1}{2} cos \frac{n\pi}{4} \right)
(Q)nC0−nC1−nC2+nC3+nC4−nC5−nC6+....^nC_0 - ^nC_1 - ^nC_2 + ^nC_3 + ^nC_4 - ^nC_5 - ^nC_6 + ....(2)2n/2(cosnπ4+sinnπ4)2^{n/2} \left( cos \frac{n\pi}{4} + sin \frac{n\pi}{4} \right)
(R)nC0+nC4+nC8+nC12+.....^nC_0 + ^nC_4 + ^nC_8 + ^nC_{12} + .....(3)2n/2(cosnπ4−sinnπ4)2^{n/2} \left( cos \frac{n\pi}{4} - sin \frac{n\pi}{4} \right)
(S)nC1+nC5+nC9+nC13+.....^nC_1 + ^nC_5 + ^nC_9 + ^nC_{13} + .....(4)2n/2(2n/2−2+12sinnπ4)2^{n/2} \left( 2^{n/2-2} + \frac{1}{2} sin \frac{n\pi}{4} \right)
(5)2n/2(2n/2−2+12cosnπ4+12sinnπ4)2^{n/2} \left( 2^{n/2-2} + \frac{1}{2} cos \frac{n\pi}{4} + \frac{1}{2} sin \frac{n\pi}{4} \right)
A

Option (2)

B

Option (3)

C

Option (1)

D

Option (4)

Answer

(P) → Option (2)

(Q) → Option (3)

(R) → Option (1)

(S) → Option (4)

Explanation

Solution

We shall show that the four sums in List I can be written in closed‐form. (The idea is to use the roots of unity filter on the binomial expansion.) For example, define

S(j)=∑k≥0(n4k+j),j=0,1,2,3.S(j)=\sum_{k\ge0} \binom{n}{4k+j},\quad j=0,1,2,3.

Then by the standard roots‐of‐unity technique

S(j)=14∑r=03ω−jr(1+ωr)n,with ω=i.S(j)=\frac{1}{4}\sum_{r=0}^{3} \omega^{-jr}(1+\omega^r)^n,\qquad\text{with }\omega=i.

Now we match the sums as follows:

  1. For (P):
    The sum is nC0+ nC1− nC2− nC3+ nC4+ nC5− ⋯^nC_0+\,^nC_1-\,^nC_2-\,^nC_3+\,^nC_4+\,^nC_5-\,\cdots which may be written as S0+S1−S2−S3.S_0+S_1-S_2-S_3. Working out the filter one finds that (P)=2n/2(cos⁡nπ4+sin⁡nπ4).(P)=2^{n/2}\Bigl(\cos\frac{n\pi}{4}+\sin\frac{n\pi}{4}\Bigr). This is exactly the expression given in Option (2).

  2. For (Q):
    The sum nC0− nC1− nC2+ nC3+ nC4− nC5− nC6+⋯^nC_0-\,^nC_1-\,^nC_2+\,^nC_3+\,^nC_4-\,^nC_5-\,^nC_6+\cdots equals S0−S1−S2+S3.S_0-S_1-S_2+S_3. A similar analysis gives (Q)=2n/2(cos⁡nπ4−sin⁡nπ4),(Q)=2^{n/2}\Bigl(\cos\frac{n\pi}{4}-\sin\frac{n\pi}{4}\Bigr), which matches Option (3).

  3. For (R):
    The sum nC0+ nC4+ nC8+ ⋯=S0,^nC_0+\,^nC_4+\,^nC_8+\,\cdots = S_0, is obtained by the roots‐of‐unity filter: (R)=14[2n+2n/2+1cos⁡nπ4]=2n−2+2n/2−1cos⁡nπ4.(R)=\frac{1}{4}\Bigl[2^n+2^{n/2+1}\cos\frac{n\pi}{4}\Bigr] =2^{n-2}+2^{n/2-1}\cos\frac{n\pi}{4}. Notice that writing 2n/2(2n/2−2+12cos⁡nπ4)=2n−2+2n/2−1cos⁡nπ4,2^{n/2}\Bigl(2^{n/2-2}+\frac{1}{2}\cos\frac{n\pi}{4}\Bigr) =2^{n-2}+2^{n/2-1}\cos\frac{n\pi}{4}, we see that (R) is given in Option (1).

  4. For (S):
    The sum nC1+ nC5+ nC9+⋯=S1,^nC_1+\,^nC_5+\,^nC_9+\cdots = S_1, is worked out to be (S)=2n−2+2n/2−1sin⁡nπ4=2n/2(2n/2−2+12sin⁡nπ4),(S)=2^{n-2}+2^{n/2-1}\sin\frac{n\pi}{4} =2^{n/2}\Bigl(2^{n/2-2}+\frac{1}{2}\sin\frac{n\pi}{4}\Bigr), which is exactly Option (4).

Hence the correct matchings are:

  • (P) → (2)
  • (Q) → (3)
  • (R) → (1)
  • (S) → (4)

Minimal Explanation

  1. Write the given sums in terms of residue class sums Sj=∑k≥0(n4k+j)S_j=\sum_{k\ge0}\binom{n}{4k+j} using the roots‐of‐unity filter.
  2. Express each sum (with the appropriate alternating signs) as a linear combination of S0,S1,S2,S3S_0, S_1, S_2, S_3.
  3. Use (1+i)n=2n/2einπ/4(1+i)^n=2^{n/2}e^{in\pi/4} and (1−i)n=2n/2e−inπ/4(1-i)^n=2^{n/2}e^{-in\pi/4} to simplify.
  4. Identify that (P) becomes 2n/2(cos⁡(nπ/4)+sin⁡(nπ/4))2^{n/2}(\cos(n\pi/4)+\sin(n\pi/4)) [Option (2)], (Q) becomes 2n/2(cos⁡(nπ/4)−sin⁡(nπ/4))2^{n/2}(\cos(n\pi/4)-\sin(n\pi/4)) [Option (3)], (R) becomes 2n/2(2n/2−2+12cos⁡(nπ/4))2^{n/2}(2^{n/2-2}+\frac{1}{2}\cos(n\pi/4)) [Option (1)], and (S) becomes 2n/2(2n/2−2+12sin⁡(nπ/4))2^{n/2}(2^{n/2-2}+\frac{1}{2}\sin(n\pi/4)) [Option (4)].