Download 1. Compute the sum of the reciprocals of 3, 5, 7, ..., 63. 2. Compute 1

Sathwik Karnik
22-Nov-2016
1. Compute the sum of the reciprocals of 3, 5, 7, ..., 63.
sum = 0;
i = 1;
Whilei ≤ 31, sum = sum +
1
2*i +1
; i ++
Print[sum]
31 674 468 729 962 723 297 623 231
18 472 920 064 106 597 929 865 025
N[sum]
1.71464
2. Compute
1
1+
.
1
1+
1
1+
1
2
1
1+
1
1
1+ 1
1+
2
5
8
5
N 
8
0.625
3. Obtain a 50 significant digit approximation to the Square Root of Pi
N
π , 50
1.7724538509055160272981674833411451827975494561224
4. What is the 1000th prime?
Prime[1000]
7919
5. Sketch the graphs of y = Sin[x], y = Sin[2x], and y=Sin[3x], 0<= x<= 2Pi on one set of axes.
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B Section
Sathwik Karnik
22-Nov-2016
B Section
Plot{Sin[x], Sin[2 * x], Sin[3 * x]}, {x, 0, 2 * Pi},
Ticks → Range0, 2 * Pi, Pi  4, AxesLabel → {"x", "y"}
y
π
4
π
π
3π
4
2
4
π
5π
3π
7π
4
2
4
x
2π
6. What is the prime factorization of 2381400?
FactorInteger[2 381 400]
{{2, 3}, {3, 5}, {5, 2}, {7, 2}}
7. Find two ways to find an approximate value of x for which 2^x = 100.
x = N[Log[100] / Log[2]]
6.64386
Reduce[2 ^ n == 100, n, Reals]
n⩵
N
2 (Log[2] + Log[5])
Log[2]
2 * (Log[2] + Log[5])

Log[2]
6.64386
8. What is the 115th Fibonacci number? The 1115th Fibonacci number?
The 115th Fibonacci number is:
Fibonacci[115]
483 162 952 612 010 163 284 885
The 1115th Fibonacci number is:
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Sathwik Karnik
22-Nov-2016
B Section
Fibonacci[1115]
46 960 625 891 577 894 920 915 085 010 622 289 470 462 518 359 149 677 075 881 383 631 822 660 )
890 718 642 869 603 700 018 836 567 361 824 279 444 479 341 088 310 462 978 732 670 769 895 389 )
845 153 583 927 059 046 832 024 176 024 794 070 671 098 298 816 588 315 827 802 770 672 734 166 )
457 585 412 100 971 385
9. What are the greatest common divisor and least common multiple of 5355 and 40425?
GCD[5355, 40 425]
105
LCM[5355, 40 425]
2 061 675
10. Find the sum of the squares of the first 20 consecutive integers.
The first method is brute force:
1 ^ 2 + 2 ^ 2 + 3 ^ 2 + 4 ^ 2 + 5 ^ 2 + 6 ^ 2 + 7 ^ 2 + 8 ^ 2 + 9 ^ 2 + 10 ^ 2 +
11 ^ 2 + 12 ^ 2 + 13 ^ 2 + 14 ^ 2 + 15 ^ 2 + 16 ^ 2 + 17 ^ 2 + 18 ^ 2 + 19 ^ 2 + 20 ^ 2
2870
The second method utilizes the classical sum formula for the sum of the first n positive square numbers:
n*(n+1)*(2*n+1)
,
6
which can be shown by induction or from the sum of the first n positive integers.
(20) * (21) * (41)
6
2870
11. Compute the sum of the reciprocals of 15, 17, 19, ..., 51.
Clear[sum, i]
sum = 0;
i = 7;
Whilei ≤ 25, sum = sum +
1
2*i +1
; i ++
Print[N[sum]]
0.6557
1
12. Compute the value of  1 +
1
2
+
1
3
1
2
+ 4 + 1 +
2
2
+
2
3
Normal way
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2
3
+ 4 + 1 +
3
2
+
3
3
3
+ 4  in at least 2 ways.
Sathwik Karnik
1
+
1
25
22-Nov-2016
1
+
2
1
+
3
1
2
+
4
+
1
2
2
+
2
+
3
2
3
+
4
1
+
3
+
2
3
+
3
B Section
3
4
2
Reorder the fractions:
1
1
25
+
2
+
1
3
+
1
1
+
2
2
2
+
3
2
+
1
3
+
2
3
+
3
3
+
1
4
+
2
4
2
6*
1
1
+
1
2
+
1
3
+
1
4
25
2
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+
3
4