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Hello friends please solve this problem number is 19.

Hello friends please solve this problem number is 19.

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Grade:12th pass

3 Answers

Deepak Kumar Shringi
askIITians Faculty 4404 Points
5 years ago
your image is not clear..im not able to see powers of x..kindly repost it
Samyak Jain
333 Points
5 years ago
?Q.  (1+x)21 + (1+x)22 + (1+x)23 + …...........+ (1+x)30
Ans. Using binomial expansion of (1 + x)n = 1 + x + x2 + …............ + xn , we can find
(1+ x)21 = 1 + x + x2 + …............... + x22
(1 + x)22 = 1 + x + x2 + ….............. + x23
...........
(1 + x)30 =1 + x + x2 + …................+ x30
\therefore  
(1+x)21 + (1+x)22 + (1+x)23 + …...........+ (1+x)30 
 = 10(1 + x + x2 + ….......... +x21) + 9x22+ 8x23 +7x23+ ....... + 2x29 + x30  .
Samyak Jain
333 Points
5 years ago
Above answer is wrong. The correct answer is follows : 
Q.  (1+x)21 + (1+x)22 + (1+x)23 + …...........+ (1+x)30 
Ans. Using binomial expansion of (1 + x)n = nC0 + nC1 x + nC2 x2 + …............ + nCn xn , we can find
(1+ x)21 = 21C0 + 21C1 x + 21C2 x2 + …............... + 21C21 x21
(1 + x)22 = 22C0 + 22C1 x + 22C2 x2 + ….............. + 22C22 x22
(1 + x)30 = 30C0 + 30C1 x + 30C2 x2 + …................+ 30C30 x30
\therefore  (1+x)21 + (1+x)22 + (1+x)23 + …...........+ (1+x)30 
= 21C0 + 21C1 x + 21C2 x2 + ….......... + 21C21 x21 + 22C0 + 22C1 x + 22C2 x2 + …......... + 22C22 x22 
  + …...................................... + 30C0 + 30C1 x + 30C2 x2 + …................+ 30C30 x30
 
 = (21C0 + 22C0 + 23C0 + ….. + 30C0) + (21C1 + 22C1 + 23C1 + ….. + 30C1) x +
  (21C2 + 22C2 + 23C2 + …..+ 30C2) x2 + …......
  …... + (28C28 + 29C28 + 30C28) x28 + (29C29 + 30C29) x29 + (30C30) x30
= (21C1 + 21C0 + 22C0 + 23C0 + ….. + 30C0 – 21C1) + (21C2 + 21C1 + 22C1 + 23C1 + ….. + 30C1 – 21C2) x + (21C3 + 21C2 + 22C2 + 23C2 + …..+ 30C2 – 21C3) x2 + …................ + (28C28 + 29C28 + 30C28) x28
 +  (29C29 + 30C29) x29 + (30C30) x30 
= (31C1 – 21C1) + (31C2 – 21C2) x + (31C3 – 21C3) x2 + …............... 
  ........+ (1 + 29 + 435) x28 + (1 + 30) x29 + (1) x30 
 = (31C1 – 21C1) + (31C2 – 21C2) x + (31C3 – 21C3) x2 + ….......... + 465 x28 + 31 x29 + x30 
You can calculate values if needed.
 Formula used : nCr + nCr+1  =  n+1Cr+1 .

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