Sequences and Series

Substitute n=1,2,3,4,5: 1·3=3; 2·4=8; 3·5=15; 4·6=24; 5·7=35.

Put n=1,2,3,4,5 into n/(n+1), keeping each numerator over the next integer.

Evaluate the powers 2¹, 2², 2³, 2⁴ and 2⁵.

Substitute n=1 through 5: the numerators are −1, 1, 3, 5 and 7, each over 6.

The factor (−1)ⁿ⁻¹ alternates the sign, while 5ⁿ⁺¹ gives 25, 125, 625, 3,125 and 15,625.

Substitute n=1 through 5: 1·6/4=3/2; 2·9/4=9/2; 3·14/4=21/2; 4·21/4=21; 5·30/4=75/2.

a_17=4(17)−3=65. a_24=4(24)−3=93.

Substitute n=7: a₇=7²/2⁷=49/128.

The sign is (−1)^8=1 and 9³=729.

Substitute n=20: a₂₀=20(20−2)/(20+3)=360/23.

Apply the recurrence repeatedly: 3(3)+2=11; 3(11)+2=35; then 107 and 323.

Start with 1,1. Each next term is the sum of the previous two: 2, then 3, then 5.

Apply the rule after the first two terms: a_3=1, a_4=0 and a_5=−1.

The first six Fibonacci terms are 1,1,2,3,5,8. Divide each next term by the current one: 1/1, 2/1, 3/2, 5/3 and 8/5.

The first term is 5/2 and the common ratio is 1/2. Therefore a_n=(5/2)(1/2)^(n−1)=5/2ⁿ. Put n=20 for a_20=5/2²⁰.

a_12=a_8r^4=192×2^4=192×16=3,072.

Let terms be ar^(n−1). Then p=ar⁴, q=ar⁷, s=ar¹⁰. Therefore q²=a²r¹⁴=ps.

Let the ratio be r. T4=−3r³ and (T2)²=9r², so r²(r+3)=0. Thus r=0 or r=−3. T7=−3r⁶, giving 0 or −2,187.

(a) The nth term is 2(√2)^(n−1). Since 128=2⁷, (√2)^(n−1)=2⁶, so n−1=12 and n=13. (b) The nth term is (√3)^n. Since 729=3⁶=(√3)^12, n=12. (c) The nth term is 3^−n; 1/19683=3^−9, so n=9.

For three consecutive G.P. terms, the middle squared equals the product of the outer terms: x²=(2/7)(7/2)=1.

This is a G.P. with a=0.15 and r=0.1. S_20=a(1−r²⁰)/(1−r)=0.15(1−10^−20)/0.9=(1/6)(1−10^−20).

The common ratio is √21/√7=√3, so a=√7 and r=√3. Apply S_n=a(r^n−1)/(r−1).

This G.P. has first term 1 and common ratio −a. Apply S_n=(1−r^n)/(1−r).

The first term is x³ and common ratio is x². Apply the finite G.P. sum formula.

Split the sum into eleven copies of 2 and the powers of 3: 22+Σ(k=1 to 11)3^k. The geometric sum is (3^12−3)/(3−1)=265,719, so the required total is 265,741.

For three terms write a/r,a,ar. Their product is a³=1, so a=1. The sum gives r+1+1/r=39/10; solving yields r=5/2 or 2/5.

S_n=3(3^n−1)/2=120 gives 3^n−1=80, so 3^n=81=3⁴ and n=4.

The next block sum is r³ times the first block, so r³=128/16=8 and r=2. Then a(1+2+4)=16, giving a=16/7.

729r⁶=64 gives r=±2/3. Sum the seven terms using S_7=a(1−r⁷)/(1−r) for each sign; the two values differ because the sign of r is not specified.

Let first term a and ratio r. a(1+r)=−4; ar⁴=4ar² implies r²=4. For r=2, a=−4/3; for r=−2, a=4.

These terms are six places apart: x=ar³,y=ar⁹,z=ar¹⁵. Then y²=a²r¹⁸=xz.

The kth term is 8(10^k−1)/9. Sum from k=1 to n: (8/9)[10(10^n−1)/9−n], which simplifies to the result.

Products are 256,128,64,32,16, a G.P. with ratio 1/2. Their sum is 256+128+64+32+16=496.

At position k the product is (ar^(k−1))(AR^(k−1))=aA(rR)^(k−1), which is a G.P.

Let the terms be a,ar,ar²,ar³. The conditions give a(r²−1)=9 and ar(1−r²)=18. Dividing yields −r=2, so r=−2 and a=3.

Write the terms as A R^(p−1), A R^(q−1), A R^(r−1). The exponents on A and R in the stated product both sum to zero.

The product is P=a^n r^(n(n−1)/2), while b=ar^(n−1). Thus (ab)^n=a^(2n)r^(n(n−1))=P².

Each term in the second block is r^n times the corresponding term in the first block. Therefore the second block sum is r^n S_n.

Write the four terms as t,tr,tr²,tr³. Both sides simplify to t⁴r²(1+r²+r⁴)²? Equivalently, factor the left as t⁴r²(1+r²+r⁴)², matching the right.

For common ratio r, 3r³=81, so r³=27 and r=3. The middle terms are 3r=9 and 3r²=27.

Set t=√(a/b)>0 and divide through by the positive powers of b. Equating the quotient to √ab reduces to (t−1)(t^(2n+1)−1)=0. For this to hold for every positive ratio t, choose 2n+1=0, so n=−1/2. (When a=b, the equality also holds for any n.)

Let u=√a,v=√b. Then u²+v²=6uv, so t=u/v satisfies t²−6t+1=0. Thus t=3±2√2. Since a/b=t², the ratio simplifies to the stated reciprocal pair.

The numbers x,y satisfy x+y=2A and xy=G². Solve t²−2At+G²=0.

After n hours the count is 30 multiplied by 2 once per hour: 30·2^n.

Amount=500(1.10)^10=500×2.59374246≈₹1,296.87.

The sum of roots is 2×8=16; their product is 5²=25. The monic quadratic is x²−16x+25=0.

By induction f(n)=3^n. The geometric sum 3+9+…+3^n=3(3^n−1)/2=120, so 3^n=81 and n=4.

5(2^n−1)=315 gives 2^n=64 and n=6. The last term is 5·2^5=160.

r²+r⁴=90. Put u=r²: u²+u−90=0=(u−9)(u+10), so u=9 and r=±3.

Let terms be a,b,c. Then a+b+c=56, b²=ac, and 2(b−7)=(a−1)+(c−21), so a+c−2b=8. Solve with a+c=56−b to get b=16, a+c=40, ac=256; hence a,c=8,32.

For 2m terms, the even-position sum is r times the odd-position sum, so total=(1+r) times odd-position sum. Set 1+r=5.

Equate the first two fractions and cross-multiply. After cancelling equal terms, 2x(b²−ac)=0, so b²=ac because x≠0. Equating the second and third fractions in the same way gives c²=bd. Thus the consecutive-term ratios agree, so a,b,c,d form a G.P.

For first term a and ratio r, R=S/(a²r^(n−1)) and P=a^n r^(n(n−1)/2). Substitute; powers cancel to give S^n.

Write a,b,c,d=t,tr,tr²,tr³. The three sums are t^n(1+r^n), t^nr^n(1+r^n), t^nr^(2n)(1+r^n), with common ratio r^n.

Their sums are 3 and 12. If the four roots are t,tr,tr²,tr³, then (c+d)/(a+b)=r²=4. Their products give q/p=r⁴=16, so the ratio is (16+1):(16−1)=17:15.

Let A/G=m/n. The roots are A±√(A²−G²). Divide by G and multiply both by n: m±√(m²−n²).

The kth term is 5(10^k−1)/9. Sum for k=1,…,n using the finite geometric sum of powers of 10.

The kth term is (2/3)(1−10^−k). Sum this expression from k=1 to n.

The kth term is (2k)(2k+2)=4k(k+1). For k=20, the term is 40×42=1,680.

The unpaid balances for interest are ₹6,000,₹5,500,…,₹500. Their sum is ₹39,000, so total interest is 12%×39,000=₹4,680. Add the ₹12,000 price.

The financed balance is ₹18,000. Interest is on balances 18,000,17,000,…,1,000; their sum is ₹171,000 and 10% is ₹17,100. Add to ₹22,000.

Each round sends four times as many letters as the previous one. The 8th set contains 4⁸=65,536 letters, so its postage is 65,536×₹0.50=₹32,768.

Simple interest per year=₹500. Add 15×500 or 20×500 to the principal.

Each year retains 80% of value. After 5 years: 15,625(0.8)^5=₹5,120.

Actual worker-days are the arithmetic sum 150+146+… over d+8 days: (d+8)[150−2(d+7)]=150d. Solving d²+15d−544=0 gives d=17 (reject negative root).