and=N+(2n 1) n-.1 +3+5+13. The identity(1+2+3++n)n 1= 13 +23 +33 + .+ n3,as the first century. Provide awas known as earlyderivation of it.14. Prove the following formula for the sum of triangularnumbers, given by the Hindu mathematicianAryabhata (circa 500):n(n +1)(n+2)t1t2 +t3 + +tnHint: Group the terms on the left-hand side in pairs,replacing t-1 + tk by k; consider the two cases wheren is odd and n is even.]15. Archimedes (287-212 B.C.) also derived the formula12+22+32 +n(n + 1)(2n + 1)+ n211for the sum of squares. Fill in any missing details in the HuSee 8-9-74ed to3714andSac 3.3 #CU9/20/19a denvahon of (1+2+3 n ++nHw 3.2# 133Pronden .(proof of it)t(+36इमई टाटterte sideiDt2 n13+23 3 n 1-1-1otos+n nn1-1-1 + 2 2-2l2-1+22+*inn+6P(I2t3+ n) n(nti):+ग(गर स) नाh(nt)(stnen tn42hCntl)(nt2)-- t+tnProve titt+t3+Hint: Group the terms in LHS in pairs, replacing(consider n is odd or even)K++Suppose n is even: (tt) (t3ttu)(tstto)tttntnJ!l

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Asked Sep 24, 2019
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Number 13 please

i sort of got most of the way

and
=N
+(2n 1) n-.
1 +3+5+
13. The identity
(1+2+3++n)
n 1
= 13 +23 +33 + .+ n3,
as the first century. Provide a
was known as early
derivation of it.
14. Prove the following formula for the sum of triangular
numbers, given by the Hindu mathematician
Aryabhata (circa 500):
n(n +1)(n+2)
t1t2 +t3 + +tn
Hint: Group the terms on the left-hand side in pairs,
replacing t-1 + tk by k; consider the two cases where
n is odd and n is even.]
15. Archimedes (287-212 B.C.) also derived the formula
12+22+32 +
n(n + 1)(2n + 1)
+ n2
11
for the sum of squares. Fill in any missing details in the
help_outline

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and =N +(2n 1) n-. 1 +3+5+ 13. The identity (1+2+3++n) n 1 = 13 +23 +33 + .+ n3, as the first century. Provide a was known as early derivation of it. 14. Prove the following formula for the sum of triangular numbers, given by the Hindu mathematician Aryabhata (circa 500): n(n +1)(n+2) t1t2 +t3 + +tn Hint: Group the terms on the left-hand side in pairs, replacing t-1 + tk by k; consider the two cases where n is odd and n is even.] 15. Archimedes (287-212 B.C.) also derived the formula 12+22+32 + n(n + 1)(2n + 1) + n2 11 for the sum of squares. Fill in any missing details in the

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Hu
See 8-9-74ed to3714
and
Sac 3.3 #
CU
9/20/19
a denvahon of (1+2+3 n ++n
Hw 3.2# 13
3
Pronde
n .(proof of it)
t
(+36
इमई टाट
terte sideiDt2 n
13+23 3 n 1-1-1
otos
+n nn
1-1-1 + 2 2-2
l2-1+22+*inn
+6
P
(I2t3+ n) n(nti)
:+ग
(गर स) ना
h(nt)
(stnen tn
4
2
hCntl)(nt2)
-- t+tn
Prove titt+t3+
Hint: Group the terms in LHS in pairs, replacing
(consider n is odd or even)
K
+
+
Suppose n is even: (tt) (t3ttu)(tstto)tttntn
J
!l
help_outline

Image Transcriptionclose

Hu See 8-9-74ed to3714 and Sac 3.3 # CU 9/20/19 a denvahon of (1+2+3 n ++n Hw 3.2# 13 3 Pronde n .(proof of it) t (+36 इमई टाट terte sideiDt2 n 13+23 3 n 1-1-1 otos +n nn 1-1-1 + 2 2-2 l2-1+22+*inn +6 P (I2t3+ n) n(nti) :+ग (गर स) ना h(nt) (stnen tn 4 2 hCntl)(nt2) -- t+tn Prove titt+t3+ Hint: Group the terms in LHS in pairs, replacing (consider n is odd or even) K + + Suppose n is even: (tt) (t3ttu)(tstto)tttntn J !l

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Expert Answer

Step 1

To prove the required identity (for all n).

Step 2

Proof by mathematic...

n 1: both sides 1;
n 2: LHS=32 9;
RHS =1+89,so true
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n 1: both sides 1; n 2: LHS=32 9; RHS =1+89,so true

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