Class 11_Chapter 9_Sequences and Series (Sum of special series) Lecture 12.pdf

#class_11 #JEE #sequences #series
Class 11 Chapter 9 Sequences and Series_ Lecture 12
#special_series
#sum_of_first_n_natural_numbers
#sum_of_squares_of_first_n_natural_numbers
#sum_of_cube_of_first_n_natural_numbers
#class_11
#class_11_math
#class_11_maths
#Mathematics_Online_Lectures
#Maths_Learning_Centre_Jalandhar

1. Dr. Pranav Sharma
Maths Learning Centre. Jalandhar.

2. Some Special Series

SUM OFFIRSTN NATURAL NUMBERS

Provetnat A+2+3 ++ +n) = En(n +1) 22, Ej le =Ím(n+D).
Consider the series (1+2+3+..+1n).

This is an arithmetic series in which a =1,d=1andl=n.

n 1 1
Sn = ZA +n) > Sy =Zn(n+ 1) Sn = 5 (a+ D}.

Hence, ear k= ¿nn +1).

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3. SUM OF THE SQUARES OF FIRST N NATURAL NUMBERS
Prove that (1? + 22 +3? +--+n?) = En(n +1)(2n +1)

Le, Eat? =in(nt+1)(Qn+1).
LES, = 127 4+27 437 +--+ n?
Consider the identity: k3 — (k — 1)? = 3k? — 3k +1. (i)
Putting k=1,2,3, (n—1),n successively in (i), we get

13-03 =3-17-3-14+1;

23-13 =3-2?-3-24+1;
33-23=3.32-3-3+1
Mm-D3-(M-2 =3.(n-1)?-3:(n-1)+1;
n-—(m-D?=3.N2-3-n+1.

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4. Adding column wise, we get
(m3 - 03) =3(12+224+324+-+n2)-3(142+3+-+n)+n
ñ

sm =3: 10-304 1) +n |a+2+34 4m =3n(0+0)]
1
= 3 i
= a(S mt) ant En + D Zn Han +)
kA
= [ak] = 2 (2n3 + 3n? +n) = in(2n? + 3n +1) =in(nt 1)(2n +1).
Henee, Digi? = Én(n +1)(2n +1).

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5. SUM OF THE CUBES OF FIRST N NATURAL NUMBER

Prove that (13 + 23 +33 + -+n?) = Ena + D} OR = Ena + D)

LaS. = 13 +23 433 tan.

Consider the identity: k* — (k - 1)? = 4k3 — 6k? +4k-1. (6)

Putting k = 1,2,3,(n—1),n successively in (i), we get
1+-0?=4:1°-6.1?+4:1=1
24-14=4-23-6-274+4-2-1
34 24=4.33-6-37+4-3-1

=

(n- 1)* — (n - 2)* -6-m-D?+4-(n-D)-1
n®-(n-1)*=4:n?-6-.n?+4:n-1.
Adding column wise, we get
(n* — 04) =4- {13 + 23 +33 +--+ 3} —6- (12 [phone] n2)
+4 -(14243+-+n)-(1+ 1+1+--to nterms)

a 4) 10)-0-) 10) +0 (4) =m

k=1 k=1 k=1

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6. 1 5 A ‘

>n*=4- ye —6-En(n+ DQ2n+1)+4-5nm+-n
k=1

CDR = Enr + 1)Qn +0 and E k = nn + 1)}

> 4. (De) =m ns Dœn +0 an nn
k=
n 5
> D e) => (n* + 2n? +n?) = i n?(n? +2n+1)= Enn+ D}
Eiai)= Ena + D} Le, (13 + 23 +33 +--+n3) = Ena + D}.

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7. SUM OF THE CUBES OF FIRST n ODD NATURAL NUMBERS
Prove that [13 + 33 + 53 + --- (2n - 1)*] = n?(2n? - 1)
Le. YB4(2k — 1)? = n2(2n? - 1).
LaS, = 19 +334 5% 4---4+ (2n —1)$.
Then, Sn = Er-ı(2k - 1)? = YR_,(8k3 — 12k? + 6k — 1)
= 8(D hark?) - 12(Uh kk?) + 604 10 — n L1+41+to nterms =n]
=8-intn +1)? ~12-En(n + Dm +1) +6 Zum + 1)-n
= 2n?(n +1)? —2n(n+1)(2n+1)+3nm+1)-n
=n(n+1)-{2n(n+1)-2(2n+1)+3}-n
= n(n + 1)(2n? —2n+1)-n=n-{(n+1)(2n? -2n+1)-1)
= n(2n? - n) = n?(2n? -1).
{13 + 33 4+ 53 +--+ (2n—1)3} = n?(2n? - 1).

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8. @ 14243 4--4+n) =3n(n+1) Le Oe) = En 1).
(i) (12 +22 432 4..+ n?) = ¿n(n +1)(2n +1)

Le, (Bike) = ¿nn + DON +1).
(i) (13 +23 +33 +--+n3) = Ent D} Le, (Chak) = Gain + vy
(6) {13 +33 +53 +--+ (2n— 1)3} = n?(2n? - 1)

Le, Di (2k - 1)3 = n?(2n? -1).

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9. Uf Si, Sz and S3 are the sums of first n natural numbers, their squares and their
cubes respectively then show that 983 = Sa(1 + 851).
We have

S¡=(1+2+3+-+)>5,=3n(n+1);
Sy = (17 +2? 43? 4+--4+n?) 3S, =in(n + 1)(2n+1);
and $3 = (17 +23 +39 42403) = =D?
983 =9 x ag Mint 1)2(2n + 1)? = nn 1)?(2n + 1)?
And, S3(1 + 85:) = In?(n +1)? -[1+8-Inn+1)}
= grin 1)?(4n? + 4n +1) = an2(n + 1)2(2n + 1)2
Hence, 983 = S$3(1+ 8S,).

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10. Find the sum ton terms of the series whose uth term isn(n +3).
we have, Ty = k(k + 3) = (k? + 3k).
sum to n terms ts given by

5,- n=) k? + 3k) = Zara)
uen DES Lien + 1) = AE DCR A) CRD) =

n(n+ 1)@n +1+9) =In(n+1)-2(n+5)=inn+1) m +5).
Hence, the required sum Is nn + 1)(n + 5).

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11. Find the sum ton terms of the series whose nth term is (2n = 1)?
we have, Ty = (2k — 1)? = (4k? — 4k +1).
sum to n terms is cane Y

= Dr } ue - 4k+1) = De D

L Maths. ringe =n]
=4-Inin+ D@n+D)-4-Inin+1)+n
2:
= zun + D(2n+1)-2n(n+1)+n
= 5 + {2n(n + 1)(2n + 1) - 6n(n + 1) + 3n)
= jinn + 1){2(2n + 1) — 6} +3n] => [4n(n + 1)(n — 1) + 3n]
SE 5 (4n? -n)= n(4n? -1).

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12. Find the sum ton terms of the series whose nth term is (n? + 2%). we have, Tr =
(12 +2").

n n

= Y ri= S02 +22 Y a

1 k=1 k=1

=|
it
a
=|
it

n(n + 1)(2n+1) + (2427 +23 4-42")

i 2
. y 2 = nn + DQn+1)
k=1

E 2(2"-1)
= nn + 1)(2n +1) + — @- 5

= Inn + 1)(2n+1)+2(2"-1).

12+2°?+..+2”isaGPl

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13. Sum the series 3 : 8 + 6 : 11 +9: 14 + ton terms.
we have Ty =(kth term of 3, 6, 9, ...) X (kth term of 2, 11, 14, ...)
= {3+ (k-1) x 3} x 18 +(k-—1) x 3} = 3k(3k + 5) = (9k? + 15k).

Sn = ) Ty = ) (9k? + 15k) 9 Ke) +15 4)
ens (¿no FDON+ 1) Yis: (¿no E 1}
= En(n + 1){(2n + 1) + 5} = 3n(n + 1)(n +3).
Hence, the required sum is 3n(n + 1)(n + 3).

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14. Sum the series1: 2+34+2°3-44+3-4-54++-ton terms.
wehave Ty =(kth term of 1, 2, 3, ...) X (kth term of 2, 3, 4, ...) x (kth term of 3,
4,5, ...)
={1+(k=1) x 1}4 (2+ (k-1) x1} (34 (k=1) x 1}
= ER E nie |

= Y ri = (eran + 21) à Jere de
ki fl
1

= qn im +1)2+3 ann + 1)(2n+1)+2 en 1)
VL = nj + D} ak =¿n(n+ DOR+D,

y pes Inn + Dye in(n + 1) {nn +4) 4 2@n 41) 49

= En(n+ 1)(n? + 5n +6) = En(n+ 1)(n + 2)(n +3).

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15. Find the sum of the series 1? + 3? + 5? + ton terms:
We have
Ty = kth term of (1? + 3? + 57 +)
= {1+ (k-1)x 2} = (2k-1)? = (4k? = 4k +1).
n n

Sn=) Te = ) (4K? - 4k +1)

k=1
=4 41? 4 01 k+(1+1+--ntímes)

1 1
=4:¿n(n+1)Qn + 1)-4:0n+1+n
=5 (2+ D(2n+1)- 6+ 1) +3) =F(4n? 1).
Hence, the required sum is (an? - 1).

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16. Find the sum 1? +(12 +2?) + (1? + 2? + 32) + tom terms.
We have
Ty, = (1? +2? +3? 4-442) = [hh ner: +1) = E +3k2 + i).

- a = E oye Durs da
k=1

k=1 k=1
171 2 1 1 Let
eg +7 zul + Dan + 1) + ee nd)

1 1 1
san +1? + TL 1)(2n +1) tort 1)

AC ED OR HDI LAB HAE (n +2).

wi.
Se
ai mil

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17. 1x2?+2x3?+....+nx(n+1)? _ 3n+5
12x24+22x3+-+n2x(n+1) 3n+1'

Prove that

we have

kth term of the wumerator = k(k + = (13 +/2k? + k) a

kth term of the denominator = k?(k + 1) = (k3 + k2).

CA AA k?)+ (The)
DETTES) Era 1)

In2(n+1)2+2-In(n+1) 2nt1)+3n(n+1)

LHS =

dn? (n+1)2+In(n+1)(2n+1)

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b-2 27 n(n+1)(n+2)(3n+5) 12 nts) _
a” GET EE 12 n(n+1)(n+2)(3n+1) (Bn+1)
RHS.
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18. 1 1
eat (5x8) (8x11)

Find the sum of n terms of the series ==

we have Ti = (kth term of 2,5,8,. Ru term of 5,8,11,...)
L

4 1 1 1
~ +(-D)x3)x(5+(k-1)x3) (Bk-DGk+2) nece ~ alas
1 1 1

kTzlar) Gkr2)) ©
Putting k= 1,2,3,n successively in (i), we get
171 1 1/1 1
ne nr) ,
SE ee en
Ta G igs Tes (GES Een

Adding column wise, we get

1
that tr Tae = — | E

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19. Find the sum of n terms of thé series 5 + 11 +194+29 +41 ++
SOLUTION LES, = 5 +11+19+29 + --+T,4at+Tp
Sn = BUF 19 +0 4+ Tyo +Ta-ı+ Ta
Ow subtraction, we get
0 = 5 +16 + 8 + 10 + 12 + ---to (n — 1)terms] -T,
(n-Dx[2x6+(n-2)x2]
2

Ty = 5+
> T, =n? 4+3n+1.
Sn = Dhar Tr = Eta? + 3k 4 1) = 12430 kin

= nn +1)@n +1) +3x5n(nF1)+n=Fn(nt2)(n44).

Ta =5+(n—-1)(n+ 4)

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20. Find the sum of the firstn terms of the series 347 +13 + 21+31+»+"
LAS, =3+74+13+21+31++T, +7,
S = 3474 13421 4-4 5 ET a +7,

On subtraction, we get 0 = 3 +14 + 6 + 8 +10 +to (n —1)terms] -T,
ST, =3 +14 +648 + to (n—1) terms] = 3 + D x [2 x 4+ (n—2) x 2]
=3+n-Dn+2)=(n+n+1).
Sn = her Tie E (1? + 41) = Ei 2 + E + mn

L1+1+1+ to N terms = nl
=n(n+ 1)(2n+ 1) +5n(n+ D+n= En(n? +3n +5).

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21. Find the 50th term of the series 2+ 3 + 6+11+18+::
LS, =24+3+6+11+18+-+T, 1 +Tp
Again Sn =2+3+6+11+-+Tp 1 +Tp.
On subtracting, we get
0=2+114+3+5+7+to(n-— 1)terms] -T,
3T, =2+114+3+5+7+to(n-— 1) terms]

We)

=2+ -2x1+(n-2)x2]=(n-1)? +2.
> Tso = {(50 — 1)? + 2} = {(49)2 + 2} = (2401 + 2) = 2403.

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22. Find the sum of the series
(33 - 23) + (53 — 45) + (73 — 65) +... to
(à n terms (öl) 10 terms.
we have Ti = {(2k + 1)? — (2k)3}

= (8k3 + 1 + 6k(2k + 1) — 8k} = (12k? + 6k +1).
() sum to n terms is given by Sn = 12 Ek? + 6% ık +n

= 12 x En(n+1)2n+1) +6x5n(n + 1) +n
=2n(n+1)(2n+1)+3n(n+1)+n
=n(n+1)[4n+2+3]+n=n[(n+1)(4n+5)+1]
= n(4n? + 9n + 6) = (4n? + 9n? + 6n).
(ii), Sum to 10 terms is given by
S10 = 10 x [4 x 10? +9 x 10 + 6] = 4960.

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23. e Bn gara? | 13423433
Find the sum of the series + De

Let Ty be the nth term of the given series. Then,

++ upto n terms.

2
ne(aterade3taacend)r_ Bi NatGe-sgärnia
M (14345+.+@n-D)) ¿(42m D) 4004
Let Sy denotes the sum of n terms of the given series. Then,

(n? + 2n+1)

dr 1
5,221; = que +2n+1)= ¿Er + 2En + £1)
1 E +1)(2n+1) 2n(n+1) }
TER + +n

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4 6 2
n
= 74 l2r" +3n+1+6n+6+6)
=, n(2n2+9n+13)
Hence, Sy = De”
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24. Find sum ton terms of the series: 1+ (24 3) + (44+54+6)+:..

Now, number of terms in first bracket is 1, in the second bracket is 2, in the third
bracket is 3, etc. Therefore, the number of terms in the uth bracket will be n.

Let the sum of the given series of n terms = S

Number of terms in S = 1+2 +34. +4n =D

Also, the first term of S is 1 and common difference is also 1.

En
la. 1+ (ED) 1] 2 (44m? + n — 2)

aan + 1)(n? +n +2)
+

s-12J

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25. Find the sum of the series

16+ 2 —1)+43 (QA — 2) +4-(1—3)+=+(n— 1) 2 Hl

also, find the coefficient of "1 in the expansion of (1 + 2x + 3x2 ++ nxt)
The rth term of the given series ts T, =r-(n=r+1) = (n+ 1)r— 72

Sum of the series Sy = DPT, = (n+ DIA, - yr? = (n+ D)Zn- Zn?

sm n(n+1) = n(n+1)(2n+1) _ n(n+1) = = _ n(n+1)(n+2)
=(n4 10% e = MEnstamianced) e laamgrar —

Now, (1+ 2x + 3x2 + + nat 1)? =
(1+ 2x 4382 ++ nx") x (1+ 2x + 3x2 + tnx)
Coefficient of al in (1 + 2x + 3x2 ++ nara)

ní(n + D(n +2)
heces Mitin 2) + ERST Anótentre Japindhar

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