Algebra> Hello friends please solve this problem n...

Hello friends please solve this problem number is 19.

shivam Prajapati , 7 Years ago

Grade 12th pass

3 Answers

Deepak Kumar Shringi

Last Activity: 7 Years ago

your image is not clear..im not able to see powers of x..kindly repost it

Samyak Jain

Last Activity: 7 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+ ….............. + x^{23
...........}

(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

Last Activity: 7 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 + ⁿC₁x + ⁿC₂ x²+ …............ + ⁿC_n 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