Series Problems in Math (Crack Puzzles with Lightning Speed in 5 Easy Steps)

This chapter focuses on some ‘Series Problems in Maths’ commonly found in competitive and entrance exams, which are designed to test one’s ability to identify patterns and predict the next term in a sequence. These problems often involve numerical or alphabetic sequences and require a logical approach to solve. By mastering these types of problems, students can improve their problem-solving skills and increase their chances of success in these exams.

Que 1: Find the value of 1¹+2²+3³+4⁴+5⁵+6⁶+7⁷+8⁸+9⁹

(a) 405071317

(b) 405071727

(c) 405071427

(d) 405071257

Solution: Here, the last three digits of the options are different, so first we will find the remainder by dividing the given expression by 8 and then match it with the remainders obtained by dividing the given options by 8. (Series Problems in Maths)

4⁴, 6⁶, 8⁸ are already divisible by 8 so remove them

The totient of 8= 4

It means, k⁴ mod 8=1, where k is not divisible by 8.

5⁵ mod 8= 5

7⁷ mod 8= 7³ mod 8=(-1)³=-1

9⁹ mod 8=1

⇒ (1+4+3+5-1+1) mod 8

⇒ 5

Now, only option (a) satisfies this condition.

⇒ 405071317 mod 8

⇒ 317 mod 8

⇒ 5

Option:(a)

Que 2: 1+12+123+1234+…….+9th term=? 

(a) 133764205

(b) 137174205

(c) 134146205

(d) 137426205

Solution: First, find the remainder of the given expression from 9

1 mod 9=1

12 mod 9=3

123 mod 9=3+3=6

1234 mod 9=6+4⇒1

12345 mod 9=1+5⇒6

123456 mod 9=6+6⇒3

1234567 mod 9=3+7⇒1

12345678 mod 9=1+8⇒0

123456789 mod 9=0+9⇒0

(1+3+6+1+6+3+1+0+0) mod 9=3 

Only options (b) and (d) satisfy the condition. So one of them is correct.

Now we have to find the last four digits of the expression. (Series Problems in Maths)

Unit Place⇒ 1+2+3+…..+9=[(9+1)/2]×9=45

Tens Place⇒ 1+2+3+…..+8=[(8+1)/2]×8=36

Hundreds Place⇒ 1+2+3+…..+7=[(7+1)/2]×7=28

Thusands Place⇒ 1+2+3+…..+6=[(6+1)/2]×6=21

Last 4 digits⇒ 21 | 28 | 36 | 45

⇒ 4205

Option:(b)

Que 3: N=1+3+6+10+15+21+……+50th term=? 

(a) 23400

(b) 22100

(c) 21780

(d) 23000

Solution: 2N=2+6+12+20+30+42+……+50th term (Series Problems in Maths)

2N=1×2+2×3+3×4+4×5+5×6+6×7+……+50×51

2N=\sum_{k=1}^{50}\limits k\times (k+1).

2N=\sum_{k=1}^{50}\limits k^2+\sum_{k=1}^{50}\limits k………(I)

\sum_{k=1}^{50}\limits=\frac{k\times (k+1)}{2}=\frac{50\times 51}{2}=25\times 51.

\sum_{k=1}^{50}\limits k^2=\frac{1}{6}k\times (k+1)\times (2k+1)=\frac{1}{6}\times 50\times 51\times 101=25\times 17\times 101.

2N=25×(51+17×101)

2N=25×(51+1717)

2N=25×1768

N=25×884= 100/4×884=22100

Option:(b)

Que 4: 3+7+13+21+31+43+57+…..+20th term=? 

(a) 3470

(b) 3380

(c) 3290

(d) 3100

Solution: N=(2²-1)+(3²-2)+(4²-3)+(5²-4)+(6²-5)+…..+(21²-20)

N=(2²+3²+4²+5²+6²+…..+21²)- (1+2+3+4+…….+20)

(1²+2²+3²+4²+5²+6²+…..+n²)=16n(n+1)(2n+1)

Put n=21 (Series Problems in Maths)

⇒3311

(1+2+3+4+…….+n)=12n(n+1)

Put n=20

⇒210

N=(3311-1)-210=3100

Option (d) is correct.

Que 5: 1+4+13+40+121+364+….+ 10th term=? 

(a) 46531

(b) 39251

(c) 44281

(d) 40511

Solution: S=1+4+13+40+121+364+…+ T_n

S=0+1+4+13+40+121+364+…T_n-1+ T_n

Subtract equation (i) from (ii):

0=1+{3+9+27+81+……+(n-1) terms}-T_n

T_n=1+\frac{3.(3^{n-1}-1)}{(3-1)}.

T_n=\frac{3^n-3+2}{2}=\frac{1}{2}\times (3^n-1).

S=\frac{1}{2}\sum 3^n-\frac{1}{2}\sum 1.

S=\frac{1}{2}\times\frac{3.(3^{n}-1)}{(3-1)}-\frac{n}{2}.

S=\frac{3^{n+1}-3-2n}{4}.

Put n=10

S=\frac{3^{11}-23}{4}=44281.

Note: 3¹¹=243²×3

Option (c) is correct.