© examsiri.com
Question : 13 of 100
Marks:
+1,
-0
Solution:
Method (1) (Proper Method)
Now
(1+x)6⋅(1+x)6=(‌6C0+‌6C1x+‌6C2x2+‌6C3x3+...+‌6C6x6)
(‌6C0+‌6C1x+‌6C2x2+‌6C3x3+...+‌6C6x6)
On comparing the coefficients of
x6 from both sides, we have
=924 Method (2) (Short-cut Method)
As, we know that,
‌nC02+‌nC12+‌nC22+‌nC32+...+‌nCn2=‌2nCn
⇒‌nC0⋅‌nC0+‌nC1⋅‌nC1+‌nC2⋅‌nC2+...+‌nCn*‌nCn=‌2nCn
⇒‌nC0⋅‌nCn+‌nC1⋅‌nCn−1+‌nC2⋅‌nCn−2+...+‌nCn‌nC0
=‌2nCn(∵‌nCr=‌nCn−r) Putting
n=6, we get
‌6C0⋅‌6C6+‌6C1⋅6C5+‌6C2⋅‌6C4+...+‌6C6⋅‌6C0=‌12C6
© examsiri.com
Go to Question: